# Working File

James Melbourne, Piotr Nayar, and Cyril Roberto

# Minimum entropy of a log-concave variable with fixed variance

James Melbourne, Piotr Nayar, and Cyril Roberto

###### Abstract.

We show that for log-concave real random variables with fixed variance the
Shannon differential entropy is minimized for an exponential random variable.
We apply this result to derive upper bounds on capacities of additive noise
channels with log-concave noise. We also improve constants in the reverse
entropy power inequalities for log-concave random variables.

###### Key words and phrases:

entropy, variance, log-concave random variables

###### 1991 Mathematics Subject Classification:

94A17 (Primary), 60E15 (Secondary)

P.N.’s research was supported by the National Science Centre, Poland, grant
2018/31/D/ST1/01355

##  1\. Introduction

For a real random variable $X$ with density $f$ its differential entropy is
defined via the formula $h(X)=h(f)=-\int f\log f$. This definition goes back
to the celebrated work of Shannon [13], but the same quantity was also
considered, without the minus sign, by physicists, including Boltzmann, in the
context of thermodynamics of gases. In fact, it is a classical fact going back
to Boltzmann [4] that under fixed variance the entropy is maximized for a
Gaussian random variable. This leads to the translation and scale invariant
inequality

$h(X)\leq\frac{1}{2}\log\operatorname{Var}(X)+\frac{1}{2}\log(2\pi e).$

One can see that in general one cannot hope for a reverse bound, since for the
density
$f_{\varepsilon}(x)=(2\varepsilon)^{-1}\mathbf{1}_{[1,1+\varepsilon]}(|x|)$
the variance stays bounded while the entropy goes to $-\infty$ as
$\varepsilon\to 0^{+}$. However, a reverse bound still holds if one imposes
some extra assumption on $X$, such as log-concavity. Recall that $X$ is said
to be log-concave if its density is of the form $f=e^{-V}$, where
$V:\mathbb{R}\to(-\infty,\infty]$ is convex. In [3] Bobkov and Madiman showed
that indeed in this class, the inequality can be reversed, up to an absolute
additive constant and it became a well-known open problem to find the optimal
such bound. The sharpest inequality to date can be found in [10] where
Marsiglietti and Kostina proved that if $X$ is log-concave, then
$h(X)\geq\frac{1}{2}\log\operatorname{Var}(X)+\log 2$. The goal of this
article is to prove the following inequality.

######  Theorem 1.1.

For a log-concave random variable $X$ we have

$h(X)\geq\frac{1}{2}\log\operatorname{Var}(X)+1$

with equality for the standard one-sided exponential random variable with
density $e^{-x}\mathbf{1}_{[0,\infty)}(x)$.

Probably the most significant generalization of entropy is the so-called Rényi
entropy of order $\alpha\in(0,\infty)\setminus\\{1\\}$, which is defined as

$h_{\alpha}(X)=h_{\alpha}(f)=\frac{1}{1-\alpha}\log\left(\int
f^{\alpha}(x)\mathrm{d}x\right),$

assuming that the integral converges, see [12]. If $\alpha\to 1$ one recovers
the usual Shannon differential entropy $h(f)=h_{1}(f)=-\int f\ln f$. Also, by
taking limits one can put $h_{0}(f)=\log|\mathrm{Supp}f|$, where
$\mathrm{Supp}f$ stand for the support of $f$ and
$h_{\infty}(f)=-\log\|f\|_{\infty}$, where $\|f\|_{\infty}$ is the essential
supremum of $f$. For $p\geq q>0$ one has

$0\leq h_{q}(f)-h_{p}(f)\leq\frac{\log q}{q-1}-\frac{\log p}{p-1},$

where the fraction $\frac{\log t}{t-1}$ is interpreted as $1$ for $t=1$, see
[5]. Thus using this bound together with Theorem 1.1 gives the following
corollary.

######  Corollary 1.2.

For $\alpha>1$ and a log-concave random variable $X$ one has

$h_{\alpha}(X)\geq\frac{1}{2}\log\operatorname{Var}(X)+\frac{\log\alpha}{\alpha-1}.$

Let us mention that the problem of minimizing Rényi entropy under fixed
variance for symmetric log-concave random variables was solved in [8] for the
case $\alpha\leq 1$ and in [2] for $\alpha>1$. Here the worst case in the
uniform random variable on an interval for $\alpha\leq\alpha^{*}$ and
symmetric exponential random variable for $\alpha>\alpha^{*}$, where
$\alpha^{*}\approx 1.241$ is the solution to the equation
$\frac{\log\alpha^{*}}{\alpha^{*}-1}=\frac{1}{2}\log 6$.

##  2\. Applications

###  2.1. Additive noise channels

We now briefly discuss an application of our main result in the context of
information theory. For more details we refer the reader to [9], where the
case of symmetric log-concave random variables was discussed. Consider the
memoryless transmission channel with power budged $P$ subject to additive
noise $N$, that is, if the input of the channel is $X$, then output produced
by the channel is $Y=X+N$, where $N$ is the noise independent of $X$.
Shannon’s celebrated channel coding theorem [13] asserts that the so-called
_capacity_ of such a channel is gives by the formula

$C_{P}(N)=\sup_{X:\ \operatorname{Var}(X)\leq P}(h(X+N)-h(N)).$

We have the following fact.

######  Proposition 2.1.

Let $N$ be a random variable with finite variance and let $Z$ be a centered
Gaussian random variable with the same variance. Then

$C_{P}(Z)\leq C_{P}(N)\leq C_{P}(Z)+D(N),$

where $D(N)=h(Z)-h(N)$ is the relative entropy of $N$ from Gaussianity.

Our Theorem 1.1 gives

$D(N)=h(Z)-h(N)\leq
h(Z)-\frac{1}{2}\log\operatorname{Var}(N)-1=\frac{1}{2}\log(2\pi
e)-1=\frac{1}{2}\log\left(\frac{2\pi}{e}\right).$

We can therefore formulate the following corollary.

######  Corollary 2.2.

Let $N$ be a log-concave noise and let $Z$ be a centered Gaussian noise with
the same variance. Then

$C_{P}(Z)\leq C_{P}(N)\leq
C_{P}(Z)+\frac{1}{2}\log\left(\frac{2\pi}{e}\right).$

It other words, using an arbitrary log-concave noise instead of the Gaussian
does not increase capacity by more than
$\frac{1}{2}\log\left(\frac{2\pi}{e}\right)<0.42$ nats.

###  2.2. Reverse EPI

Recall that the entropy power of a random variable $X$ is defined by
$\mathcal{{N}}(X)=\frac{1}{2\pi e}\exp(2h(N))$. Note that for a Gaussian
random variable one has $\mathcal{{N}}(Z)=\operatorname{Var}(Z)$ and in
general $\operatorname{Var}(X)=\operatorname{Var}(Z)$ implies
$\mathcal{{N}}(X)\leq\mathcal{{N}}(Z)$. Theorem 1.1 can be rewritten in the
form $\mathcal{{N}}(X)\geq\frac{e}{2\pi}\operatorname{Var}(X)$. The entropy
power inequality of Shannon and Stam [13, 14] states that for independent
random variables $X,Y$ one has

$\mathcal{{N}}(X+Y)\geq\mathcal{{N}}(X)+\mathcal{{N}}(Y).$

It is of interest to obtain reverse bounds. Under log-concavity assumption the
authors of [10] showed that if $X,Y$ are uncorellated, then
$\mathcal{{N}}(X+Y)\leq\frac{\pi e}{2}(\mathcal{{N}}(X)+\mathcal{{N}}(Y))$.
Using Theorem 1.1 we can improve this result.

######  Corollary 2.3.

Let $X,Y$ be log-concave uncorrelated real random variables. Then

$\mathcal{{N}}(X+Y)\leq\frac{2\pi}{e}\left(\mathcal{{N}}(X)+\mathcal{{N}}(Y)\right).$

Indeed, one has

$\mathcal{{N}}(X+Y)\leq\operatorname{Var}(X+Y)=\operatorname{Var}(X)+\operatorname{Var}(Y)\leq\frac{2\pi}{e}\left(\mathcal{{N}}(X)+\mathcal{{N}}(Y)\right).$

We can also define the Rényi entropy power via
$\mathcal{{N}}_{\alpha}(X)=\exp(2h_{\alpha}(X))$, where we removed the
normalizing constant $2\pi e$ for simplicity. Similarly as in [2], Theorem 2
from [7] together with our Theorem 1.1 gives the inequality

(1) $C_{-}(\alpha)\operatorname{Var}(X)\leq N_{\alpha}(X)\leq
C_{+}(\alpha)\operatorname{Var}(X),\qquad\alpha>1$

where

$C_{-}(\alpha)=\alpha^{\frac{2}{\alpha-1}},\qquad\qquad
C_{+}(\alpha)=\frac{3\alpha-1}{\alpha-1}\left(\frac{2\alpha}{3\alpha-1}\right)^{\frac{2}{1-\alpha}}B\left(\frac{1}{2},\frac{\alpha}{\alpha-1}\right)^{2}.$

Here $B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$ stands for the Beta
function. The right inequality does not need log-concavity. Thus, using the
same computation as for the case $\alpha=1$ we get the following corollary.

######  Corollary 2.4.

If $X,Y$ are log-concave uncorrelated real random variables, then for
$\alpha>1$ one has

$\mathcal{{N}}_{\alpha}(X+Y)\leq\frac{C_{+}(\alpha)}{C_{-}(\alpha)}\left(\mathcal{{N}}_{\alpha}(X)+\mathcal{{N}}_{\alpha}(Y)\right).$

##  3\. Reductions

###  3.1. Decreasing Rearrangement

######  Definition 3.1 (Decreasing Rearrangement).

For a measurable set $A\subseteq\mathbb{R}$ let $|A|$ denote its Lebesgue
measure and let us define

$A^{\downarrow}=(0,|A|)$

with the interpretation of $(0,0)$ as the empty set.

######  Definition 3.2.

For a measurable function $f:\mathbb{R}\to[0,\infty]$, define
$f^{\downarrow}:(0,\infty)\to[0,\infty]$

$\displaystyle
f^{\downarrow}(x)=\int_{0}^{\infty}\mathbbm{1}_{\\{y:f(y)>\lambda\\}^{\downarrow}}(x)d\lambda.$

We observe that $f^{\downarrow}$ is fully characterized by the equality
$\\{f>\lambda\\}^{\downarrow}=\\{f^{\downarrow}>\lambda\\}$ and in particular,
equimeasurable (with respect to the Lebesgue measure) functions possess
identical decreasing rearrangements.

######  Proposition 3.3.

For $f:\mathbb{R}\to[0,\infty)$, its rearrangement $f^{\downarrow}$ satisfies,

(2) $\displaystyle
f^{\downarrow}(x)=\sup\\{\lambda:x\in\\{f>\lambda\\}^{\downarrow}\\}$

and

(3) $\displaystyle\\{f^{\downarrow}>\lambda\\}=\\{f>\lambda\\}^{\downarrow}.$

###### Proof.

Representation (2) follows directly from the definition of $f^{\downarrow}$.
To prove (3) observe that by (2) the condition $f^{\downarrow}(x)>\lambda$ is
equivalent to the existence of $\lambda^{\prime}>\lambda$ such that
$x\in\\{f>\lambda^{\prime}\\}^{\downarrow}$. It is therefore enough to prove
the following equivalence

$\exists_{\lambda^{\prime}>\lambda}\ \
x\in\\{f>\lambda^{\prime}\\}^{\downarrow}\quad\iff\quad
x\in\\{f>\lambda\\}^{\downarrow}.$

By the nestedness the implication $\implies$ is trivial. To show the other
direction assume that $x\in\\{f>\lambda\\}^{\downarrow}$. This is equivalent
to $x<|\\{f>\lambda\\}|$. By the continuity of Lebesgue measure

$x<|\\{f>\lambda\\}|=\left|\bigcup_{n}\left\\{f>\lambda+\frac{1}{n}\right\\}\right|=\lim_{n}|\\{f>\lambda+1/n\\}|.$

By taking $\lambda^{\prime}=\lambda+\frac{1}{n}$ for large enough $n$ we get
$\lambda^{\prime}>\lambda$ such that
$x\in\\{f>\lambda^{\prime}\\}^{\downarrow}$. ∎

######  Proposition 3.4.

For $\varphi$ measurable and $f$ non-negative,

$\int\varphi(f(x))dx=\int\varphi(f^{\downarrow}(x))dx.$

###### Proof.

This is equivalent to the statement that $f$ pushes the Lebesgue measure $dm$
forward to the same measure $f^{\downarrow}$ pushes the Lebesgue measure to.
Thus it suffices to check

$f_{\\#}m(\lambda,\infty)=f_{\\#}^{\downarrow}m(\lambda,\infty).$

This is just $|\\{f>\lambda\\}|=|\\{f^{\downarrow}>\lambda\\}|$, which follows
from the characterization
$\\{f^{\downarrow}>\lambda\\}=\\{f>\lambda\\}^{\downarrow}$. ∎

Note that taking $\varphi(x)=-x\log x$ shows that the decreasing rearrangement
$f^{\downarrow}$ preserves the entropy of a density function $f$.

######  Definition 3.5.

For a random variable $X$ with density function $f$, define $X^{\downarrow}$
to be a random variable drawn from the density function $f^{\downarrow}$.

When $X$ has density $f$ we will write the variance as ${\rm Var}(X)$ and
${\rm Var}(f)$ interchangeably.

######  Lemma 3.6.

For non-negative $X$, and increasing, non-negative $\varphi$,

$\mathbb{E}\varphi(X^{\downarrow})\leq\mathbb{E}\varphi(X).$

###### Proof.

The proof follows from the elementary observation that for $a\geq 0$,

(4) $\displaystyle\left|(a,\infty)\cap A\right|\geq\left|(a,\infty)\cap
A^{\downarrow}\right|.$

Using the layer-cake decomposition,

$\displaystyle\mathbb{E}\varphi(X)$
$\displaystyle=\int_{0}^{\infty}\int_{0}^{\infty}\left(\int_{0}^{\infty}\mathbbm{1}_{\\{\varphi>\lambda\\}}(x)\mathbbm{1}_{\\{f>\sigma\\}}(x)dx\right)d\lambda
d\sigma$
$\displaystyle=\int_{0}^{\infty}\int_{0}^{\infty}\left|\\{\varphi>\lambda\\}\cap\\{f>\sigma\\}\right|d\lambda
d\sigma$
$\displaystyle\geq\int_{0}^{\infty}\int_{0}^{\infty}\left|\\{\varphi>\lambda\\}\cap\\{f^{\downarrow}>\sigma\\}\right|d\lambda
d\sigma=\mathbb{E}\varphi(X^{\downarrow}),$

where the inequality follows from (4). ∎

######  Lemma 3.7.

For probability densities $f_{0},f_{1}$ and $\lambda\in[0,1]$ the density
$f=\lambda f_{0}+(1-\lambda)f_{1}$ has variance,

${\rm Var}(f)=\lambda{\rm Var}(f_{0})+(1-\lambda){\rm
Var}(f_{1})+\lambda(1-\lambda)(\mu_{1}-\mu_{0})^{2}$

where $\mu_{i}$ denotes the barycenter of $f_{i}$, that is $\mu_{i}=\int
xf_{i}(x)dx$.

###### Proof.

$\displaystyle{\rm Var}(f)=$ $\displaystyle\int x^{2}(\lambda
f_{0}(x)+(1-\lambda)f_{1}(x))dx-\left(\int x(\lambda
f_{0}(x)+(1-\lambda)f_{1}(x))dx\right)^{2}$ $\displaystyle=$
$\displaystyle\lambda\left(\int x^{2}f_{0}(x)dx-\left(\int
xf_{0}(x)dx\right)^{2}\right)+(1-\lambda)\left(\int x^{2}f_{1}(x)dx-\left(\int
xf_{1}(x)dx\right)^{2}\right)$ $\displaystyle+\lambda\left(\int
xf_{0}(x)dx\right)^{2}+(1-\lambda)\left(\int xf_{1}(x)dx\right)^{2}-\left(\int
x(\lambda f_{0}(x)+(1-\lambda)f_{1}(x))dx\right)^{2}$ $\displaystyle=$
$\displaystyle\lambda{\rm Var}(f_{0})+(1-\lambda){\rm
Var}(f_{1})+\lambda(1-\lambda)(\mu_{1}-\mu_{0})^{2}.$

∎

######  Theorem 3.8.

For $X$ log-concave,

${\rm Var}(X)\leq{\rm Var}(X^{\downarrow}).$

###### Proof.

We prove the result when $X$ has a density $f$ given by a unimodal step
function by induction, that is
$f=\sum_{k=0}^{n}\lambda_{k}\mathbbm{1}_{I_{k}}/|I_{k}|$ with $I_{k}$
intervals satisfying $I_{k+1}\subsetneq I_{k}$ and $\lambda_{k}>0$. An easy
limiting argument gives the result for log-concave $X$. When $n=0$, $X$ is
uniform and the result is immediate. Assuming the result for $n^{\prime}<n$,
we proceed. The density of $f$ can be written as $\lambda
f_{0}+(1-\lambda)f_{1}$, with $\lambda=\lambda_{0}$,
$f_{0}=\frac{\mathbbm{1}_{I_{0}}}{|I_{0}|}$, and
$f_{1}=\sum_{k=1}^{n}\frac{\lambda_{k}}{1-\lambda_{0}}\frac{\mathbbm{1}_{I_{k}}}{|I_{k}|}$.
Observing that $f_{1}$ takes strictly less values than $f$ and that by affine
invariance of the inequality we may assume that $I_{0}=(0,1)$, and by
considering $\tilde{X}=1-X$, we may assume without loss of generality, that
$\int xf_{1}(x)dx\leq\frac{1}{2}$.

By Lemma 3.7,

${\rm Var}(f)=\lambda{\rm Var}(f_{0})+(1-\lambda){\rm
Var}(f_{1})+\lambda(1-\lambda)\left(\frac{1}{2}-\int xf_{1}(x)dx\right)^{2}.$

Observe that $f^{\downarrow}=\lambda
f_{0}^{\downarrow}+(1-\lambda)f_{1}^{\downarrow}$. Applying Lemma 3.7 to
$f^{\downarrow}$,

${\rm Var}(f^{\downarrow})=\lambda{\rm
Var}(f_{0}^{\downarrow})+(1-\lambda){\rm
Var}(f_{1}^{\downarrow})+\lambda(1-\lambda)\left(\frac{1}{2}-\int
xf_{1}^{\downarrow}(x)dx\right)^{2}.$

Clearly $f_{0}^{\downarrow}=f_{0}$. By induction ${\rm Var}(f_{1})\leq{\rm
Var}(f_{1}^{\downarrow})$ and by Lemma 3.6, applied to $\varphi(x)=x$, we have
$\frac{1}{2}\geq\int xf_{1}(x)dx\geq\int xf_{1}^{\downarrow}(x)dx$. The result
follows. ∎

######  Lemma 3.9.

For $X$ log-concave,

$\frac{e^{2h(X)}}{{\rm Var}(X)}\geq\frac{e^{2h(X^{\downarrow})}}{{\rm
Var}(X^{\downarrow})}$

This follows immediately from Theorem 3.8 and Proposition 3.4 (with
$\varphi(x)=-x\log x$) since decreasing rearrangement will preserve entropy
and increase variance. The following result, a special case of a result from
[11], will thus allow us to reduce our problem to $X$ log-concave with
monotone decreasing density.

######  Proposition 3.10 ([11] Proposition 7.5.8).

For $X$ log-concave, $X^{\downarrow}$ is log-concave as well.

###### Proof.

Note that $X$ has a log-concave density $f$ if and only if

(5)
$\displaystyle(1-t)\\{f>\lambda_{1}\\}+t\\{f>\lambda_{2}\\}\subseteq\\{f>\lambda_{1}^{1-t}\lambda_{2}^{t}\\}.$

Thus, to show that $X^{\downarrow}$ is log-concave it suffices to prove

$(1-t)\\{f>\lambda_{1}\\}^{\downarrow}+t\\{f>\lambda_{2}\\}^{\downarrow}\subseteq\\{f>\lambda_{1}^{1-t}\lambda_{2}^{t}\\}^{\downarrow}$

But since both sets are open intervals it suffices to prove that the right
hand side has bigger volume, which follows Brunn-Minkowski inequality and the
set theoretic inclusion in (5),

$\displaystyle|(1-t)\\{f>\lambda_{1}\\}^{\downarrow}+t\\{f>\lambda_{2}\\}^{\downarrow}|$
$\displaystyle=|(1-t)\\{f>\lambda_{1}\\}+t\\{f>\lambda_{2}\\}|$
$\displaystyle\leq|\\{f>\lambda_{1}^{1-t}\lambda_{2}^{t}\\}|$
$\displaystyle=|\\{f>\lambda_{1}^{1-t}\lambda_{2}^{t}\\}^{\downarrow}|.$

∎

###  3.2. Degrees of freedom

Our goal is to show that in order to prove Theorem 1.1 it suffices to consider
functions of the form $e^{-V}$ where $V$ is a two-piece affine on a finite
interval. This is a standard argument appearing e.g. in [8], so we only sketch
it.

_Step 1._ Let $\sigma^{2}=\operatorname{Var}(X)$. By the previous subsection,
in order to prove the inequality $h(f)\geq\log(e\sigma)$ it suffices to
consider non-increasing densities on $[0,\infty)$. In fact by an approximation
argument we can assume that the support of $f$ is a finite interval. Indeed if
we define $f_{n}=e^{-V}\mathbbm{1}_{(0,n)}/c_{n}$, where
$c_{n}=\int_{0}^{n}e^{-V}$ is the normalizing constant, then clearly $c_{n}\to
1$ as $n\to\infty$ and by the Lebesgue dominated convergence theorem
$\operatorname{Var}(f_{n})\to\operatorname{Var}(f)$. We also have

$h(f_{n})=\frac{1}{c_{n}}\int_{0}^{n}Ve^{-V}+\frac{\log
c_{n}}{c_{n}}\int_{0}^{n}e^{-V}\to h(f)$

by the Lebesgue dominated convergence theorem, as
$|V|e^{-V}=|V|e^{-V/2}e^{-V/2}\leq e^{-V/2}$ is integrable.

_Step 2._ We can therefore fix the interval $[0,L]$ on which the function $f$
is defined. Let $A=A_{L,\sigma}$ denote the set of log-concave non-increasing
densities supported in $[0,L]$ and having variance $\sigma^{2}$. We have
$f(0)^{2}\operatorname{Var}(X)\leq f(0)^{2}\mathbb{E}X^{2}\leq 2$, see [1]. To
see this we follow the argument from [8]. By scaling we can assume that
$f(0)=1$. If $g(x)=e^{-x}\mathbbm{1}_{[0,\infty)}(x)$ then by log-concavity of
$f$ the function $f-g$ changes sign in exactly one point $a>0$ and thus

$\mathbb{E}X^{2}-2=\int_{0}^{\infty}x^{2}(f(x)-g(x))\mathrm{d}x=\int_{0}^{\infty}(x^{2}-a^{2})(f(x)-g(x))\mathrm{d}x\leq
0$

since the integrant is non-positive. This shows that $f$ is bounded by
$2\sigma^{-2}$ and in particular $h(f)=-\int f\log f\geq-\log f(0)$ is bounded
from below. This shows that the quantity $M=\inf\\{h(f):f\in A_{L,\sigma}\\}$
is finite. Let $f_{n}$ be such that $h(f_{n})\to M$. By a straightforward
adaptation of Lemma 12 from [8] we get that $(f_{n})$ has a convergent
subsequence $f_{n_{k}}\to f_{0}\in A$ and by the Lebesgue dominated
convergence theorem $h(f_{0})=M$. This shows that the infimum of $h(f)$ is
attained on $A$.

_Step 3._ We now apply the theory of degrees of freedom due to Fradelizi and
Guédon from [6]. We say that $f\in A$ has $d$ degrees of freedom in there
exist $\varepsilon>0$ and linearly independent functions $f_{1},\ldots,f_{d}$
such that $f_{\delta}:=f+\sum_{i=1}^{d}\delta_{i}f_{i}$ is log-concave and
non-incresing for all $|\delta_{i}|\leq\varepsilon$. Suppose $f$ has at least
$4$ degrees of freedom. Then

$\int f_{\delta}(x)\mathrm{d}x=\int f(x)\mathrm{d}x,\qquad\int
xf_{\delta}(x)\mathrm{d}x=\int xf(x)\mathrm{d}x,\qquad\int
x^{2}f_{\delta}(x)\mathrm{d}x=\int x^{2}f(x)\mathrm{d}x$

is a system of $3$ linear equations in variables
$\delta_{1},\ldots,\delta_{4}$ and thus has nontrivial linear subspace of
solutions. Thus there is $\delta$ such that $f_{\delta},f_{-\delta}\in A$ and

$h(f)=h(\frac{1}{2}f_{\delta}+\frac{1}{2}f_{-\delta})>\frac{1}{2}h(f_{\delta})+\frac{1}{2}h(f_{-\delta})$

since entropy is a strictly convex functional. Thus $f$ is not the extremizer
of $h(f)$. This shows that extremizers have to have at most $3$ degrees of
freedom and Step IV of the proof of Theorem 1 from [8] shows that such
functions must be piecewise log-affine with at most two pieces.

###  3.3. Localization

Alternatively, we can proceed with extreme point analysis following Fradelizi
and Guedon [6]. For a fixed compact interval $K\subseteq\mathbb{R}$, and upper
semi-continuous functions $g_{1},g_{2}$ define $\mathcal{P}_{g}$ to be the
space of log-concave probability measures $\mu$ supported in $K$ such that

$\int g_{i}d\mu\geq 0.$

We will use the following special case of Fradelizi and Guedon.

######  Theorem 3.11 ([6] Theorem 1).

Let $\nu$ be an extreme point of the convex hull of $\mathcal{P}_{g}$, then
$\nu$ is a point mass, or $\nu$ has density $e^{-V}$ with respect to the
Lebesgue measure where on the support of $\nu$,
$V=\max\\{\varphi_{1},\varphi_{2}\\}$ for $\varphi_{i}$ affine.

When a density $f$ has the form $e^{-V}$ for
$V=\max\\{\varphi_{1},\varphi_{2}\\}$ for $\varphi_{i}$ affine, we will say
that $f$ is two piece log-affine.

######  Lemma 3.12.

For $X$ log-concave on $\mathbb{R}$,

$\frac{e^{2h(X)}}{{\rm Var}(X)}\geq\inf_{Z\in\mathcal{K}}\frac{e^{2h(Z)}}{{\rm
Var}(Z)},$

where $\mathcal{K}$ is the space of compactly supported log-concave variables
with monotone density on this support of the form
$f=e^{-\max\\{\varphi_{0},\varphi_{1}\\}}$ for $\varphi_{i}$ affine.

###### Proof.

Recalling the truncation argument it suffices to consider $X\sim\mu$ with
density, supported on $[0,L]$ for some $L>0$. Fix $X$ and take

$\displaystyle g_{1}(x)$ $\displaystyle=\mathbb{E}[X]-x$ $\displaystyle
g_{2}(x)$ $\displaystyle=x^{2}-\mathbb{E}[X^{2}].$

For $Z\sim\nu$ an extreme point of $\mathcal{P}_{g}$, since $Z$ is non-
negative by $\int g_{i}d\nu\geq 0$, we have

${\rm Var}(Z)\geq{\rm Var}(X)>0,$

so $\nu$ is not a point mass and hence by Theorem 3.11, $Z$ has density of the
form $f=e^{-\max\\{\varphi_{1},\varphi_{2}\\}}$. Since
$X\sim\mu\in\mathcal{P}_{g}$ by definition, if we let
$\mathcal{E}(\mathcal{P}_{f})$ denote the extreme points of the convex hull of
$\mathcal{P}_{f}$, by the Krein-Milman $\mu$ belongs to the closure of the
convex hull of $\mathcal{E}(\mathcal{P}_{f})$. Now let us show that this
implies that

(6) $\displaystyle
h(X)\geq\inf_{Z\sim\nu\in\mathcal{E}(\mathcal{P}_{g})}h(Z).$

Indeed the entropy is concave and upper semi-continuous in the weak topology
on when restricted to compact sets (as can be seen from the more well known
fact that the relative entropy is lower semicontinuous, and on compact sets
$h(X)=h(U)-D(X||U)$ where $U$ is the uniform distribution on the compact set),
thus writing $\mu$ as $\lim_{n}\mu_{n}$ for a sequence of $\mu_{n}$ that can
be expressed as convex combination of extreme points, that is
$\mu_{n}=\sum_{i=1}^{k_{n}}\lambda_{n}(i)\nu_{n}(i)$ for
$\nu_{n}(i)\in\mathcal{E}(\mathcal{P}_{g})$, we have

$\displaystyle
h(\mu)\geq\limsup_{n}h(\mu_{n})=\limsup_{n}h\left(\sum_{i=1}^{k_{n}}\lambda_{n}(i)\nu_{n}(i)\right)\geq\limsup_{n}\sum_{i=1}^{k_{n}}\lambda_{n}(i)h(\nu_{n}(i))\geq\inf_{\nu\in\mathcal{E}(\mathcal{P}_{g})}h(\nu).$

Since every element of $\mathcal{P}_{g}$ has variance no smaller than $X$, it
follows that

$\frac{e^{2h(X)}}{{\rm
Var}(X)}\geq\inf_{Z\sim\nu\in\mathcal{E}(\mathcal{P}_{g})}\frac{e^{2h(Z)}}{{\rm
Var}(Z)}.$

Consider $Z^{\downarrow}$ for $Z\in\mathcal{E}(\mathcal{P}_{g})$, and applying
Lemma 3.9, we have

$\frac{e^{2h(X)}}{{\rm
Var}(X)}\geq\inf_{Z\sim\nu\in\mathcal{E}(\mathcal{P}_{g})}\frac{e^{2h(Z^{\downarrow})}}{{\rm
Var}(Z^{\downarrow})}$

Direct computation, shows that if $Z$ has a two-piece log-affine density, then
$Z^{\downarrow}$ does as well, completing the proof. ∎

##  4\. The scheme of the computation

###  4.1. Three-point inequality

From the previous section we can assume that $f=e^{-V}$, where $V$ is two-
piece affine and non-decreasing on an interval $[0,L]$. In fact by scale
invariance we can assume that we have the following parametrization of our
function

$g(t)=e^{-\frac{t}{a}}\mathbbm{1}_{[-ax,0]}(t)+e^{-\frac{t}{b}}\mathbbm{1}_{[0,-yb]}(t),\qquad
f=\frac{g}{c},\qquad c=\int g=a(e^{x}-1)-b(e^{y}-1),$

where

$a\geq b>0,\qquad x\geq 0,\qquad y\leq 0.$

Then $f$ is a probability density and we have

$\int
xg(x)\mathrm{d}x=a^{2}\left(e^{x}(1-x)-1\right)-b^{2}\left(e^{y}(1-y)-1\right)$

and

$\int
x^{2}g(x)\mathrm{d}x=a^{3}\left(e^{x}\left(x^{2}-2x+2\right)-2\right)-b^{3}\left(e^{y}\left(y^{2}-2y+2\right)-2\right).$

Also

$-\int g\log g=a\left(e^{x}(1-x)-1\right)-b(e^{y}(1-y)-1).$

We want to prove the inequality

$-\int f\log f\geq\frac{1}{2}\log\operatorname{Var}(f)+1.$

This is

$e^{-2\int f\log f}\geq e^{2}\operatorname{Var}(f).$

In terms of $g$

$e^{-2}e^{-2\int\frac{g}{c}\log\frac{g}{c}}\geq\frac{1}{c}\int
x^{2}g(x)\mathrm{d}x-\frac{1}{c^{2}}\left(\int xg(x)\mathrm{d}x\right)^{2}.$

We have

$e^{-2}e^{-2\int\frac{g}{c}\log\frac{g}{c}}=e^{-2}e^{-\frac{2}{c}\int g\log
g+2\log c}=e^{-2}c^{2}e^{-\frac{2}{c}\int g\log g},$

so after multiplying both sides by $c^{2}$ we want to prove

$e^{-2}c^{4}e^{-\frac{2}{c}\int g\log g}\geq c\int
x^{2}g(x)\mathrm{d}x-\left(\int xg(x)\mathrm{d}x\right)^{2}.$

Observe that

$e^{-2}e^{-\frac{2}{c}\int g\log
g}=\exp\left(2\frac{a\left(e^{x}(1-x)-1\right)-b(e^{y}(1-y)-1)}{a(e^{x}-1)-b(e^{y}-1)}-2\right)=\exp\left(2\frac{-axe^{x}+bye^{y}}{a(e^{x}-1)-b(e^{y}-1)}\right).$

Our goal is therefore to prove

$c^{4}\exp\left(2\frac{-axe^{x}+bye^{y}}{a(e^{x}-1)-b(e^{y}-1)}\right)\geq
c\int x^{2}g(x)\mathrm{d}x-\left(\int xg(x)\mathrm{d}x\right)^{2}.$

Equivalently

$\displaystyle\left(a(e^{x}-1)-b(e^{y}-1)\right)^{4}\exp\left(-2\frac{ae^{x}x-be^{y}y}{a(e^{x}-1)-b(e^{y}-1)}\right)$
$\displaystyle\qquad\geq\left(a(e^{x}-1)-b(e^{y}-1)\right)\left(a^{3}\left(e^{x}\left(x^{2}-2x+2\right)-2\right)-b^{3}\left(e^{y}\left(y^{2}-2y+2\right)-2\right)\right)$
$\displaystyle\qquad\qquad-\left(a^{2}\left(e^{x}(1-x)-1\right)-b^{2}\left(e^{y}(1-y)-1\right)\right)^{2}.$

The inequality is invariant under $(a,b)\to(ta,tb)$ and therefore we can
assume that $b=1$ and $a\geq 1$. Let us define the function

$\displaystyle G(c,x,y)=$
$\displaystyle\left(c(e^{x}-1)-(e^{y}-1)\right)^{4}\exp\left(-2\frac{ce^{x}x-e^{y}y}{c(e^{x}-1)-(e^{y}-1)}\right)$
$\displaystyle\qquad-\left(c(e^{x}-1)-(e^{y}-1)\right)\left(c^{3}\left(e^{x}\left(x^{2}-2x+2\right)-2\right)-\left(e^{y}\left(y^{2}-2y+2\right)-2\right)\right)$
$\displaystyle\qquad\qquad+\left(c^{2}\left(e^{x}(1-x)-1\right)-\left(e^{y}(1-y)-1\right)\right)^{2}.$

Our goal is to prove that $G(c,x,y)\geq 0$ for $x\geq 0$, $y\leq 0$ and $c\geq
1$.

###  4.2. Fifth derivative

The first crucial observation of the proof is that
$\frac{\partial^{5}G}{\partial c^{5}}$ has a sign. We have the following
lemma.

######  Lemma 4.1.

For a positive integer $n$ and some real numbers $A,B,C,D$ let
$h_{n}(t)=(At+B)^{n-1}e^{\frac{Ct+D}{At+B}}$. Then

$h_{n}^{(n)}(t)=\frac{(BC-AD)^{n}}{(At+B)^{n+1}}e^{\frac{Ct+D}{At+B}}.$

###### Proof.

Taylor expanding the exponent and using Leibniz rule under the sum we get

$\displaystyle h_{n}^{(n)}(t)$
$\displaystyle=\left(\sum_{k=0}^{\infty}\frac{1}{k!}(Ct+D)^{k}(At+B)^{n-1-k}\right)^{(n)}=\left(\sum_{k=n}^{\infty}\frac{1}{k!}(Ct+D)^{k}(At+B)^{n-1-k}\right)^{(n)}$
$\displaystyle=\sum_{k=n}^{\infty}\frac{1}{k!}\sum_{j=0}^{n}{n\choose
j}C^{j}A^{n-j}(Ct+D)^{k-j}(At+B)^{j-k-1}\frac{k!}{(k-j)!}\cdot\frac{(k-j)!}{(k-n)!}(-1)^{n-j}$
$\displaystyle=\sum_{j=0}^{n}{n\choose
j}C^{j}A^{n-j}(-1)^{n-j}\sum_{k=0}^{\infty}(Ct+D)^{k+n-j}(At+B)^{j-k-n-1}$
$\displaystyle=\sum_{j=0}^{n}{n\choose
j}C^{j}A^{n-j}(-1)^{n-j}(Ct+D)^{n-j}(At+B)^{j-n-1}e^{\frac{Ct+D}{At+B}}$
$\displaystyle=e^{\frac{Ct+D}{At+B}}(At+B)^{-(n+1)}\sum_{j=0}^{n}{n\choose
j}[C(At+B)]^{j}[-A(Ct+D)]^{n-j}$
$\displaystyle=e^{\frac{Ct+D}{At+B}}(At+B)^{-(n+1)}[C(At+B)-A(Ct+D)]^{n}$
$\displaystyle=e^{\frac{Ct+D}{At+B}}(At+B)^{-(n+1)}(CB-AD)^{n}.$

∎

Applying this lemma with

$A=e^{x}-1,\quad B=-(e^{y}-1),\quad C=-2xe^{x},\quad D=2ye^{y}$

we get

$\frac{\partial^{5}G}{\partial
c^{5}}=\frac{32\left(e^{x+y}(x-y)-e^{x}x+e^{y}y\right)^{5}e^{-2\frac{ce^{x}x-e^{y}y}{c\left(e^{x}-1\right)-(e^{y}-1)}}}{\left(c\left(e^{x}-1\right)-(e^{y}-1)\right)^{6}}.$

To prove that this is non-positive one has to show that
$e^{x}x-e^{y}y-e^{x+y}(x-y)\geq 0$. For $x=0$ we have equality, thus it is
enough to show positivity of the derivative in $x$, namely
$e^{x}(x+1)-e^{x+y}(x-y+1)\geq 0$ for $x\geq 0$. Equivalently
$x+1-e^{y}(x-y+1)\geq 0$. For $y=0$ we have equality and the derivative in $y$
is $-e^{y}(x-y)$, which is clearly non-positive. Thus the inequality holds for
$y\leq 0$.

###  4.3. From $G^{(5)}$ to $G$

By $G^{(j)}$ we denote the $j$th derivative of $G(c,x,y)$ in $c$. We claim
that in order to show that $G(c,x,y)\geq 0$ it is enough to prove the
following inequalities

$G^{(4)}(\infty,x,y)\geq 0,\quad G^{(3)}(1,x,y)\geq 0,\quad G^{(2)}(1,x,y)\geq
0,\quad G^{(1)}(1,x,y)\geq 0,\quad G(1,x,y)\geq 0.$

Indeed, the first inequality together with $G^{(5)}(c,x,y)\leq 0$ implies that
$G^{(4)}(c,x,y)\geq 0$. This together with $G^{(3)}(1,x,y)\geq 0$ implies that
$G^{(3)}(c,x,y)\geq 0$. Repeating the argument based on the above list of
inequalities finishes the argument. The rest of the proof is a verification of
these inequalities.

###  4.4. Crucial technical bound

The inequalities $G^{(j)}(1,x,y)\geq 0$ can always be written in the form

(7) $e^{2\frac{e^{y}y-e^{x}x}{e^{x}-e^{y}}}A_{j}(x)-B_{j}(x)\geq 0,$

where $A_{j}$ and $B_{j}$ are polynomials in $x,y,e^{x},e^{y}$. The
problematic exponential factor makes the inequalities intractable in the
current form. Therefore, one needs to bound the exponent. The following lemma
provides a very tight bound on this expression.

######  Lemma 4.2.

For every $x,y\in\mathbb{R}$ we have

$\left(\left(\frac{e^{2}}{6}-1\right)(x-y)^{2}e^{x+y}+(e^{x}-e^{y})^{2}+e^{2+x+y}\right)e^{-2\frac{e^{x}x-e^{y}y}{e^{x}-e^{y}}}\geq
1$

In the proof we shall need the following lemma, see [2].

######  Lemma 4.3.

Suppose the sequence of coefficients of the series
$\sum_{n=0}^{\infty}a_{n}x^{n}$ has sign pattern $(+,-)$. Then $f(x)=0$ for
exactly one point $x>0$.

###### Proof.

Let $f(x)=a_{0}+a_{1}x+\ldots+a_{k}x^{k}-b_{k+1}x^{k+1}+\ldots$, where
$a_{i},b_{i}$ are non-negative. Then $f(x)=0$ is equivalent with

$\frac{a_{0}}{x^{k}}+\frac{a_{1}}{x^{k-1}}+\ldots+a_{k}=b_{k+1}x+b_{k+2}x^{2}+\ldots$

The left hand side is decreasing and the right hand side is increasing. ∎

###### Proof of Lemma 4.2.

This inequality is invariant under $(x,y)\to(x+t,y+t)$ and therefore it is
enough to consider $y=-x$ in which case we get

$\left(\left(\frac{2e^{2}}{3}-4\right)x^{2}+4\sinh^{2}(x)+e^{2}\right)e^{-2x\coth(x)}\geq
1$

Let us define

$f(x)=\left(\left(\frac{2e^{2}}{3}-4\right)x^{2}+4\sinh^{2}(x)+e^{2}\right)e^{-2x\coth(x)}$

It is enough to prove the inequality for $x\geq 0$ since the function is even.
We will show that $f^{\prime}$ on $\mathbb{R}_{+}$ has sign pattern $(+,-)$.
This together with $f(0)=\lim_{x\to\infty}f(x)=1$ will finish the proof.

We have

$e^{2x\coth(x)}f^{\prime}(x)=\frac{2}{3}\left(\frac{x\left(-12x^{2}+2e^{2}\left(x^{2}+1\right)+e^{2}\cosh(2x)\right)}{\sinh^{2}(x)}+\frac{\left(12x^{2}-e^{2}\left(2x^{2}+3\right)\right)\cosh(x)}{\sinh(x)}\right).$

We therefore have to examine the sign pattern of

$x\left(-12x^{2}+2e^{2}\left(x^{2}+1\right)+e^{2}\cosh(2x)\right)+\left(12x^{2}-e^{2}\left(2x^{2}+3\right)\right)\cosh(x)\sinh(x).$

This is

$x\left(-12x^{2}+2e^{2}\left(x^{2}+1\right)+e^{2}\cosh(2x)\right)+\frac{1}{2}\left(12x^{2}-e^{2}\left(2x^{2}+3\right)\right)\sinh(2x)$

Taking $t=2x$ gives

$\frac{1}{2}\left(3t^{2}-e^{2}\left(\frac{t^{2}}{2}+3\right)\right)\sinh(t)+\frac{1}{2}t\left(-3t^{2}+2e^{2}\left(\frac{t^{2}}{4}+1\right)+e^{2}\cosh(t)\right)$

This is an odd function whose odd Taylor coefficient in front of $t^{n}$ for
$n\geq 5$ is equal to

$-\frac{2e^{2}(n-2)!(3(n-1)!-n!)+\left(e^{2}-6\right)(n-1)!n!}{4(n-2)!(n-1)!n!}=-\frac{2e^{2}(3-n)+(e^{2}-6)n(n-1)}{4n!},$

while the lower coefficients vanish. We have to show that the sequence

$c_{n}=2e^{2}(3-n)+(e^{2}-6)n(n-1)$

has sign pattern $(-,+)$. Since $c_{5}=-1.7751<0$ this follows from the fact
that $c_{n}$ is a convex quadratic function of $n$.

∎

######  Remark.

The rest of the proof highly relies on symbolic computational software
(Mathematica) in order to generate relevant formulas without making mistakes.
We shall mention the Mathematica commands that we use in the crucial
computations.

In the case of the first, second and third derivative the proof has the
following structure. Let

$C(x,y)=\left(\frac{e^{2}}{6}-1\right)(x-y)^{2}e^{x+y}+(e^{x}-e^{y})^{2}+e^{2+x+y}$

be the function from Lemma 4.2. After ensuring that $A_{i}(x,y)\geq 0$, we
estimate the exponent using the above bound. This leads to the inequality

$A_{i}(x,y)+C(x,y)B_{i}(x,y)\geq 0.$

We then expand the left hand side in $x$ and show that the coefficients
$P_{n}(y)$ in front of $\frac{x^{n}}{n!}$ are non-negative, which finishes the
proof. In the main body of the paper we only show the nonnegativity of
$P_{n}(y)$ for $n\geq n_{0}$, where $n_{0}$ is some explicit small number. The
verification of $P_{n}(y)$ for $n<n_{0}$ is a straightforward algorithmic (but
sometimes a bit lengthy) exercise, we therefore move this part to the
appendix.

##  5\. Nonnegativity of $G(1,x,y)$ and $G^{(4)}(\infty,x,y)$

###  5.1. The function

We are going to show the inequality

$G(1,x,y)=e^{-2\frac{e^{x}x-e^{y}y}{e^{x}-e^{y}}}\left(e^{x}-e^{y}\right)^{4}-(e^{x}-e^{y})^{2}+e^{x+y}(x-y)^{2}\geq
0.$

The inequality is invariant under the scaling $(x,y)\to(x+t,y+t)$. This is
understendable, since the case $c=1$ corresponds to only one slope, so we gain
the freedom of shifting. Therefore we can assume that $x+y=0$. In this case
one has to prove

$e^{-2x\frac{e^{x}+e^{-x}}{e^{x}-e^{-x}}}\left(e^{x}-e^{-x}\right)^{4}-(e^{x}-e^{-x})^{2}+4x^{2}\geq
0.$

This can be rewritten as

$f(x):=4e^{-2x\coth x}\frac{\sinh^{4}x}{\sinh^{2}x-x^{2}}\geq 1.$

It is enough to consider only $x>0$ since the expression is even. The left
hand side converges to $1$ as $x\to\infty$ and therefore it is enough to show
that the function is decreasing. We have

$f^{\prime}(x)=-\frac{8x\sinh^{2}(x)e^{-2x\coth(x)}\left(x^{2}-2\sinh^{2}(x)+x\sinh(x)\cosh(x)\right)}{\left(x^{2}-\sinh^{2}(x)\right)^{2}}.$

It is therefore enough to show the inequality

$x^{2}+x\sinh(x)\cosh(x)-2\sinh^{2}(x)\geq 0.$

This is

$2x^{2}+x\sinh(2x)-4\sinh^{2}(x)\geq 0.$

It is enough to show that the derivative is non-negative for $x\geq 0$, that
is

$4x+\sinh(2x)+2x\cosh(2x)-4\sinh(2x)\geq 0.$

Therefore for $t\geq 0$ one needs to show

$g(t):=2t+t\cosh(t)-3\sinh(t)\geq 0.$

Since $g(0)=g^{\prime}(0)=g^{\prime\prime}(0)$, it is enough to observe that
$g^{\prime\prime\prime}(t)=t\sinh(t)\geq 0$.

###  5.2. The fourth derivative

Using the Mathematica command

Limit[D[G[c, x, y], c, c, c, c], c $\to$ Infinity]

and simplifying further the expression shows that

$G^{(4)}(\infty,x,y)=24\left(e^{-\frac{2e^{x}x}{e^{x}-1}}\left(e^{x}-1\right)^{4}+e^{x}x^{2}-(e^{x}-1)^{2}\right).$

We are going to prove that this is nonnegative. We first observe that

$(e^{x}-1)^{2}-e^{x}x^{2}\geq 0,\qquad x\geq 0$

The coefficient in front of $\frac{x^{n}}{n!}$ is $2^{n}-n(n-1)-2>0$ for
$n\geq 4$, whereas the lower coefficients vanish. Putting the term
$(e^{x}-1)^{2}-e^{x}x^{2}$ to the right hand side and taking the logarithm
shows that it is enough to prove

$f(x):=-2\frac{e^{x}x}{e^{x}-1}+4\log(e^{x}-1)-\log((e^{x}-1)^{2}-e^{x}x^{2})>0.$

The limit of the left hand side as $x\to\infty$ is equal to $0$. Therefore it
is enough to show that the function is decreasing. We have

$(e^{x}-1)^{2}f^{\prime}(x)=-\frac{e^{x}x\left(2e^{x}\left(x^{2}+4\right)+e^{2x}(x-4)-x-4\right)}{(e^{x}-1)^{2}-e^{x}x^{2}}.$

It is therefore enough to show that

$h(x):=2e^{x}\left(x^{2}+4\right)+e^{2x}(x-4)-x-4\geq 0.$

First five derivatives of $h$ at $x=0$ vanish. We have

$h^{\prime\prime}(x)=2e^{x}\left(x^{2}+4x+2e^{x}(x-3)+6\right)$

It is therefore enough to show that

$g(x):=x^{2}+4x+2e^{x}(x-3)+6\geq 0.$

First three derivatives of $g$ at $0$ vanish, so it is enough to observe that
$g^{\prime\prime\prime}(x)=2e^{x}x\geq 0$.

##  6\. First derivative

We have

$G^{(1)}(1,x,y)=e^{\frac{2e^{y}y-2e^{x}x}{e^{x}-e^{y}}}A_{1}(x)+B_{1}(x)$

where

$A_{1}(x,y)=2\left(e^{x}-e^{y}\right)^{2}\left(e^{x+y}(x-y-2)-e^{x}(x+2)+2e^{2x}+e^{y}(y+2)\right)$

and

$B_{1}(x,y)=e^{x+y}\left(3x^{2}-2x(2y+1)+y^{2}+2y+4\right)+e^{x}\left(x^{2}+2x+4\right)-4e^{2x}-e^{y}\left(y^{2}+2y+4\right).$

###  6.1. The factor $A_{1}$ is non-negative

We now claim that the factor

$A(x,y)=e^{x+y}(x-y-2)-e^{x}(x+2)+2e^{2x}+e^{y}(y+2)$

is non-negative and therefore $A_{1}(x,y)\geq 0$. We have $A(0,y)=0$ and

$e^{-x}\frac{\partial A}{\partial x}=e^{y}(x-y-1)-x+4e^{x}-3.$

It is enough to show that this expression is non-negative. For $x=0$ we get
$1-e^{y}(y+1)\geq 0$ since for $y<-1$ there is nothing to prove and for
$y\in[-1,0]$ we have $e^{y}(y+1)\leq y+1\leq 1$. Thus it is enough to see that

$\frac{\partial}{\partial x}(e^{y}(x-y-1)-x+4e^{x}-3)=e^{y}-1+4e^{x}\geq
4e^{x}-1\geq 3>0.$

Using Lemma 4.2 we are left with proving

$A_{1}(x,y)+C(x,y)B_{1}(x,y)\geq 0.$

###  6.2. Taylor expansion

Now, the left hand side $L(x,y)$ can be expanded in $x$ and the coefficient
R[n,y] can be found using Mathematica command

$R[n\textunderscore,y\textunderscore]:=\textrm{SeriesCoefficient}[L[x,y],\\{x,0,n\\}].$

After manually multiplying by $n!$ (obtaining P[n,y]) and using the command

$\textrm{Collect}[P[n,y],\\{e^{y},y,3^{n},2^{n},n\\}]$

to collect terms one gets that the coefficient in front of $\frac{x^{n}}{n!}$
is of the form

$P_{n}(y)=a_{0}+e^{y}(b_{0}+b_{1}y+b_{2}y^{2})+e^{2y}(c_{0}+c_{1}y+c_{2}y^{2}+c_{3}y^{3}+c_{4}y^{4})+e^{3y}(d_{0}+d_{1}y+y^{2}),$

where for $n\geq 4$ we have

$a_{0}(n)=3^{n-2}(n-1)n,\qquad d_{0}=3n(n-1),\qquad d_{1}=-4n,\qquad$
$\displaystyle b_{0}(n)$
$\displaystyle=3^{n-3}\left(\left(21-2e^{2}\right)n^{2}+\left(2e^{2}-21\right)n-108e^{2}\right)$
$\displaystyle\qquad+\frac{1}{27}2^{n-5}\left(\left(9e^{2}-54\right)n^{4}+\left(108-18e^{2}\right)n^{3}+\left(351e^{2}-1242\right)n^{2}+\left(1188+522e^{2}\right)n+3456e^{2}\right)$
$\displaystyle b_{1}(n)$
$\displaystyle=4\left(e^{2}-9\right)3^{n-2}n-\frac{1}{3}\left(e^{2}-6\right)2^{n-3}n\left(n^{2}+n+14\right)$
$\displaystyle b_{2}(n)$
$\displaystyle=\frac{1}{3}2^{n-3}\left(\left(e^{2}-6\right)n^{2}+\left(3e^{2}-18\right)n+16e^{2}-120\right)+\left(15-2e^{2}\right)3^{n-1}$

and

$\displaystyle c_{0}(n)$
$\displaystyle=\left(\left(5-\frac{2e^{2}}{3}\right)n^{2}+\left(\frac{2e^{2}}{3}-5\right)n-4e^{2}\right)$
$\displaystyle\qquad+\frac{1}{3}2^{n-5}\left(\left(3e^{2}-18\right)n^{4}+\left(132-22e^{2}\right)n^{3}+\left(133e^{2}-510\right)n^{2}+\left(396-210e^{2}\right)n+384e^{2}\right)$
$\displaystyle c_{1}(n)$
$\displaystyle=\left(2-\frac{e^{2}}{3}\right)n^{2}+\left(\frac{5e^{2}}{3}-10\right)n-2e^{2}$
$\displaystyle\qquad+\frac{1}{3}2^{n-3}\left(\left(30-5e^{2}\right)n^{3}+\left(21e^{2}-126\right)n^{2}+\left(288-80e^{2}\right)n+48e^{2}\right)$
$\displaystyle c_{2}(n)$
$\displaystyle=2^{n}\left(\left(\frac{e^{2}}{2}-3\right)n^{2}+\left(6-e^{2}\right)n+\frac{5e^{2}}{3}-6\right)+\left(\left(1-\frac{e^{2}}{6}\right)n^{2}+\left(\frac{5e^{2}}{6}-5\right)n-\frac{5e^{2}}{3}+6\right)$
$\displaystyle c_{3}(n)$
$\displaystyle=-\frac{1}{6}\left(e^{2}-6\right)\left(2^{n}(3n-2)-2n+2\right)$
$\displaystyle c_{4}(n)$
$\displaystyle=\frac{1}{6}\left(e^{2}-6\right)\left(2^{n}-1\right).$

###  6.3. Signs of the coefficients

We first examine the signs of the coefficients of $P_{n}$ for $n\geq 4$.
Clearly

$a_{0}(n)>0,\qquad d_{0}(n)>0,\qquad d_{1}(n)<0,\qquad c_{3}(n)<0,\qquad
c_{4}(n)>0,\qquad b_{1}(n)<0.$

We also have

$\displaystyle c_{2}(n)$ $\displaystyle\geq
2^{n}\left(\frac{1}{2}n^{2}-\frac{3}{2}n+6\right)+(-7+n+\frac{1}{4}n^{2})\geq
16\left(\frac{1}{2}n^{2}-\frac{3}{2}n+6\right)+\left(-7+n-\frac{1}{4}n^{2}\right)$
$\displaystyle\geq 7n^{2}-23n+89>0.$

Also

$\displaystyle c_{1}(n)$
$\displaystyle\leq\left(-14+\frac{5}{2}n-\frac{2}{5}n^{2}\right)+2^{n}\left(15-12n+\frac{3}{2}n^{2}-\frac{1}{4}n^{3}\right)$
$\displaystyle\leq\left(-14+\frac{5}{2}n-\frac{2}{5}n^{2}\right)+16\left(15-12n+\frac{3}{2}n^{2}-\frac{1}{4}n^{3}\right)$
$\displaystyle=-4n^{3}+\frac{118n^{2}}{5}-\frac{379n}{2}+226=\frac{1}{5}n^{2}\left(118-20n\right)-\frac{379n}{2}+226.$

This is obviously negative for $n\geq 6$. The case $n=4,5$ is checked
directly. We also have

$\displaystyle c_{0}(n)$
$\displaystyle\geq\frac{7n^{2}}{100}+2^{n}\left(\frac{n^{4}}{25}-\frac{8n^{3}}{25}+\frac{49n^{2}}{10}-13n+29\right)-\frac{2n}{25}-30$
$\displaystyle\geq\frac{7n^{2}}{100}+16\left(\frac{n^{4}}{25}-\frac{8n^{3}}{25}+\frac{49n^{2}}{10}-13n+29\right)-\frac{2n}{25}-30$
$\displaystyle=\frac{1}{100}\left(64n^{4}-512n^{3}+7847n^{2}-20808n+43400\right).$

We have

$64n^{4}-512n^{3}+7847n^{2}-20808n+43400=64n^{3}(n-8)+n(7847n-20808)+43400.$

which is clearly positive for $n\geq 8$. The cases $n=4,5,6,7$ are checked
directly.

We have

$b_{2}(n)\geq
2^{n}\left(\frac{n^{2}}{20}+\frac{17n}{100}-\frac{2}{25}\right)+\frac{7}{100}3^{n}>0.$

Summarizing, we get

$c_{2}(n)>0,\quad c_{1}(n)<0,\quad c_{0}(n)>0,\quad b_{2}(n)>0.$

The only coefficient which may not have the correct sign is $b_{0}(n)$. We
have

$b_{0}(n)\geq
2^{n}\left(\frac{n^{4}}{100}-\frac{3n^{3}}{100}+\frac{3n^{2}}{2}+5n+29\right)+3^{n}\left(\frac{23}{100}n(n-1)-30\right).$

The first term is positive. The second term is positive for $n\geq 12$. In
fact also $b_{0}(11)>0$. We therefore have

$b_{0}(n)>0,\ n\geq 11,\qquad b_{0}(n)<0,\ 4\leq n\leq 10.$

###  6.4. Positivity of $P_{n}(y)$ for $n\geq 6$

Note that for $n\geq 11$ all the coefficients have correct signs, which
implies $P_{n}(y)\geq 0$.

For $n=9,10$ one has $P_{n}(y)\geq a_{0}(n)+e^{y}b_{0}(n)\geq
a_{0}(n)+b_{0}(n)>0$. For $n=6,7,8$ we keep more terms

$P_{n}(y)\geq a_{0}(n)+e^{y}b_{0}(n)+e^{2y}c_{0}(n).$

We now have to show that for $t\in[0,1]$ one has
$a_{0}(n)+tb_{0}(n)+t^{2}c_{0}(n)\geq 0$. It is enough to observe that the
discriminant $b_{0}(n)^{2}-4a_{0}(n)c_{0}(n)$ is negative.

Positivity of $P_{n}(y)$ for $1\leq n\leq 5$ is proved in the Appendix. We
discourage the reader from checking these computations, since they are almost
algorithmic. We provide a graph of these functions of readers convenience.

Figure 1. The graphs of
$7P_{1}(y),P_{2}(y),\frac{1}{3!}P_{3}(y),\frac{1}{4!}P_{4}(y)$ and
$\frac{1}{5!}P_{5}(y)$.

##  7\. Second derivative

Using the command

$\textrm{Simplify}[D[G[c,x,y],c,c]/.c\to 1]$

and further simplifying the expressions for $A_{2}$ and $B_{2}$ we find

$G^{(2)}(1,x,y)=e^{-2\frac{e^{x}x-e^{y}y}{e^{x}-e^{y}}}A_{2}(x,y)+B_{2}(x,y),$

where

$\displaystyle A_{2}(x,y)=$ $\displaystyle
12\left(e^{x}-1\right)^{2}\left(e^{x}-e^{y}\right)^{2}+12\left(e^{x}-1\right)\left(e^{x+y}(x-y)-e^{x}x+e^{y}y\right)\left(e^{x}-e^{y}\right)$
$\displaystyle\qquad+4\left(e^{x+y}(x-y)-e^{x}x+e^{y}y\right)^{2}$

and

$B_{2}(x,y)=e^{x+y}\left(6x^{2}-4x(y+2)+4(y+2)\right)+2e^{x}\left(3x^{2}+4x+8\right)-12e^{2x}-4e^{y}(y+2)-4.$

###  7.1. The factor $A_{2}$ is non-negative

We shall show $A_{2}(x,y)>0$ by showing that the first two terms sum up to
something positive. This sum is of the form
$12\left(e^{x}-1\right)\left(e^{x}-e^{y}\right)R(x,y)$, where

(8) $R(x,y)=e^{x+y}(x-y-1)-e^{x}(x+1)+e^{2x}+e^{y}(y+1).$

We have $R(0,y)=0$ and using $e^{x}\geq 1+x$

$\frac{\partial R}{\partial
x}=e^{x}\left(e^{y}(x-y)+2\left(e^{x}-1\right)-x\right)\geq
e^{x}\left(2\left(e^{x}-1\right)-x\right)\geq e^{x}x\geq 0.$

###  7.2. Taylor expansion

We expand the expression

$A_{2}(x,y)+C(x,y)B_{2}(x,y)$

in $x$. The coefficient in front of $\frac{1}{n!}x^{n}$ is

$P_{n}(y)=a_{0}+e^{y}(b_{0}+b_{1}y+b_{2}y^{2})+e^{2y}(c_{0}+c_{1}y+c_{2}y^{2}+c_{3}y^{3})+e^{3y}(d_{0}+d_{1}y),$

where for $n\geq 4$ one has

$a_{0}(n)=\frac{2}{3}3^{n}\left(n^{2}-3n-12\right)+2^{n}\left(n^{2}+5n+8\right)$

and

$\displaystyle d_{0}(n)$ $\displaystyle=6n^{2}-14n+8$ $\displaystyle d_{1}(n)$
$\displaystyle=-4(n-1)$ $\displaystyle b_{0}(n)$
$\displaystyle=\frac{1}{3}\left(\left(12-2e^{2}\right)n^{2}+\left(2e^{2}-48\right)n-12e^{2}-48\right)+3^{n-2}\left(\left(18-2e^{2}\right)n^{2}+\left(2e^{2}-6\right)n-108e^{2}+72\right)$
$\displaystyle\qquad+\frac{1}{3}2^{n-4}\left(\left(3e^{2}-18\right)n^{4}+\left(60-10e^{2}\right)n^{3}+\left(113e^{2}-486\right)n^{2}+\left(60+86e^{2}\right)n+768e^{2}+384\right)$
$\displaystyle b_{1}(n)$
$\displaystyle=\frac{1}{3}2^{n-2}\left(-3\left(e^{2}-6\right)n^{3}+\left(e^{2}-6\right)n^{2}-6\left(5e^{2}-38\right)n+240\right)+\frac{4}{3}\left(\left(e^{2}-12\right)n-9\right)$
$\displaystyle\qquad+\frac{4}{3}\ 3^{n}\left(\left(e^{2}-7\right)n-6\right)$
$\displaystyle b_{2}(n)$
$\displaystyle=\frac{1}{3}\left(e^{2}-6\right)2^{n-2}\left(3n^{2}+5n+32\right)+2\left(6-e^{2}\right)3^{n}+\left(4-\frac{2e^{2}}{3}\right)$

and moreover

$\displaystyle c_{0}(n)$
$\displaystyle=\frac{1}{3}2^{n-4}\left(\left(3e^{2}-18\right)n^{4}+\left(156-26e^{2}\right)n^{3}+\left(145e^{2}-534\right)n^{2}+\left(492-314e^{2}\right)n+384e^{2}-768\right)$
$\displaystyle\qquad+\left(14-\frac{4e^{2}}{3}\right)n^{2}+\left(6+\frac{4e^{2}}{3}\right)n-8e^{2}+8$
$\displaystyle c_{1}(n)$
$\displaystyle=\frac{1}{3}2^{n-1}\left(\left(12-2e^{2}\right)n^{3}+\left(11e^{2}-66\right)n^{2}+\left(102-29e^{2}\right)n+24e^{2}+24\right)$
$\displaystyle\qquad+\left(4-\frac{2e^{2}}{3}\right)n^{2}+\left(\frac{10e^{2}}{3}-12\right)n-4e^{2}-16$
$\displaystyle c_{2}(n)$
$\displaystyle=\frac{1}{3}2^{n-2}\left(\left(7e^{2}-42\right)n^{2}+\left(138-23e^{2}\right)n+16e^{2}-48\right)+\left(\frac{4e^{2}}{3}-8\right)n-\frac{4e^{2}}{3}$
$\displaystyle c_{3}(n)$
$\displaystyle=\frac{1}{3}\left(6-e^{2}\right)2^{n}(n-2)-\frac{2e^{2}}{3}+4$

We have $P_{1}\equiv 0$.

###  7.3. Signs of the coefficients

It is straightforward to see that

$a_{0}(n)>0,\qquad d_{0}(n)>0,\qquad d_{1}(n)<0,\qquad c_{3}(n)<0.$

We have

$c_{2}(n)\geq
2^{n}\left(0.8n^{2}-3n+5\right)+n-10>16\left(0.8n^{2}-3n+5\right)+n-10>0.$

We now claim that $c_{1}(n)<0$. We have

$c_{1}(n)<-0.9n^{2}+2^{n}\left(-0.4n^{3}+2.6n^{2}-18n+34\right)+13n-45.$

It is easy to prove that $-0.4n^{3}+2.6n^{2}-18n+34<0$ for $n\geq 4$ and thus
we can bound $2^{n}\geq 16$ and get

$c_{1}(n)<-6.4n^{3}+40.7n^{2}-275n+499<0.$

Finally let us show that $c_{0}(n)>0$. We have

$c_{0}(n)\geq(4n^{2}+15n-52)+2^{n}\left(0.08n^{4}-n^{3}+11n^{2}-39n+43\right)>0.$

since both terms are positive for $n\geq 4$.

We shall now show that $b_{0}(n)>0$ for $n\geq 12$. We can assume that $n\geq
21$, since the remaining cases can be checked directly. Bounding crudely

$\displaystyle b_{0}(n)$
$\displaystyle\geq-2^{n}n^{3}+3^{n}\left(0.3n^{2}-81\right)-n^{2}-12n-46\geq-2^{n}n^{3}+3^{n}\left(0.3n^{2}-81\right)-3n^{2}$
$\displaystyle\geq-2^{n}n^{3}+\frac{1}{9}3^{n}n^{2}-3n^{2}\geq-2^{n}n^{3}+\frac{1}{10}3^{n}n^{2}=\frac{1}{10}n^{2}2^{n}\left(\left(\frac{3}{2}\right)^{n}-10n\right)>0.$

###  7.4. Positivity of $P_{n}(y)$ for $n\geq 7$

From the signs of $c_{i}$ and $d_{i}$ we see that

$P_{n}(y)\geq a_{0}+e^{y}(b_{0}+b_{1}y+b_{2}y^{2}).$

We are going to show that

(9) $a_{0}+e^{y}(b_{0}+b_{1}y+b_{2}y^{2})\geq 0\qquad\textrm{for}\quad n\geq
7.$

We first assume that $8\leq n\leq 11$, in which case we write

$a_{0}+e^{y}(b_{0}+b_{1}y+b_{2}y^{2})\geq
a_{0}-|b_{0}|-e^{-1}|b_{1}|-\frac{1}{4}e^{-2}|b_{2}|>0.$

where the last inequality can be verified directly. Now assume that $n\geq
12$. In this case $b_{0}(n)>0$ and we have

$a_{0}+e^{y}(b_{0}+b_{1}y+b_{2}y^{2})\geq a_{0}+e^{y}(b_{1}y+b_{2}y^{2})\geq
a_{0}-e^{-1}|b_{1}|-\frac{1}{4}e^{-2}|b_{2}|>0.$

Observe that

$|b_{1}(n)|\leq 2^{n}\left(n^{3}+n^{2}+n+20\right)+3^{n}n+7n+12\leq 2^{n}\cdot
4n^{3}+3^{n}n+2^{n}n^{3}=2^{n}\cdot 5n^{3}+3^{n}n$

and

$|b_{2}(n)|\leq 2^{n}\left(0.35n^{2}+0.6n+4\right)+3\cdot 3^{n}+1\leq
2^{n}n^{2}+3\cdot 3^{n}.$

Therefore

$e^{-1}|b_{1}|+\frac{1}{4}e^{-2}|b_{2}|\leq 2^{n}\cdot 2n^{3}+3^{n}n.$

We also have

$a_{0}(n)\geq\frac{2}{3}3^{n}(n^{2}-3n-12)\geq 3^{n-1}n^{2}.$

It is therefore enough to show that for $n\geq 12$ one has

$3^{n-1}n^{2}\geq 2^{n}\cdot 2n^{3}+3^{n}n.$

Note that

$3^{n-1}n^{2}-3^{n}n\geq\frac{1}{4}n^{2}3^{n}\geq 2^{n}\cdot 2n^{3},$

since $8n<(\frac{3}{2})^{n}$ for $n\geq 12$.

Let us now prove (9) for $n=7$. One has

$1+e^{y}\left(\frac{b_{0}}{a_{0}}+\frac{b_{1}}{a_{0}}y+\frac{b_{2}}{a_{0}}y^{2}\right)\geq
1+e^{y}\left(-1-\frac{3}{5}y-\frac{1}{10}y^{2}\right).$

Since the right hand side vanishes for $y=0$, it is enough to show that it is
decreasing. This is true since the derivative is equal to
$-\frac{1}{10}e^{y}(y+4)^{2}$.

We have $P_{0}\equiv P_{1}\equiv 0$. Positivity of $P_{n}(y)$ for $2\leq n\leq
5$ is proved in the Appendix. We provide a graph of these functions of readers
convenience.

Figure 2. The graphs of
$P_{2}(y),\frac{1}{3!}P_{3}(y),\frac{1}{4!}P_{4}(y),\frac{1}{5!}P_{5}(y)$ and
$\frac{1}{6!}P_{6}(y)$.

##  8\. Third derivative

Using the command

$\textrm{Simplify}[D[G[c,x,y],c,c,c]/.c\to 1]$

and further simplifying the expressions, we find that

$G^{(3)}(1,x,y)=e^{-2\frac{e^{x}x-e^{y}y}{e^{x}-e^{y}}}\left(S(x,y)+24\frac{(e^{x}-1)T(x,y)^{2}}{e^{x}-e^{y}}-8\frac{T(x,y)^{3}}{(e^{x}-e^{y})^{2}}\right)+U(x,y),$

where

$\displaystyle S(x,y)$
$\displaystyle=12\left(e^{x}-1\right)^{2}\left(e^{x+y}(3x-3y-2)-e^{x}(3x+2)+2e^{2x}+e^{y}(3y+2)\right)$
$\displaystyle T(x,y)$ $\displaystyle=e^{x}x-e^{y}y-e^{x+y}(x-y)$
$\displaystyle U(x,y)$
$\displaystyle=-6\left(\left(x^{2}-2x+2\right)\left(-e^{x+y}\right)-e^{x}\left(3x^{2}+2x+6\right)+4e^{2x}+2e^{y}+2\right).$

###  8.1. Preparatory bound

Our first step is to show that

$\frac{(e^{x}-1)T(x,y)^{2}}{e^{x}-e^{y}}\geq\frac{T(x,y)^{3}}{(e^{x}-e^{y})^{2}}.$

This is equivalent to

$(e^{x}-1)(e^{x}-e^{y})\geq T(x,y),$

which simplifies to

$R(x,y)=e^{x+y}(x-y-1)-e^{x}(x+1)+e^{2x}+e^{y}(y+1)\geq 0.$

The function $R$ appeared in (8) and the inequality $R(x,y)\geq 0$ was already
verified.

It is therefore enough to show that

$e^{-2\frac{e^{x}x-e^{y}y}{e^{x}-e^{y}}}\left(S(x,y)+16\frac{(e^{x}-1)T(x,y)^{2}}{e^{x}-e^{y}}\right)+U(x,y)\geq
0$

or equivalently

$e^{-2\frac{e^{x}x-e^{y}y}{e^{x}-e^{y}}}A_{3}(x,y)+B_{3}(x,y)\geq 0$

with

$A_{3}(x,y)=S(x,y)(e^{x}-e^{y})+16(e^{x}-1)T(x,y)^{2},\qquad
B_{3}(x,y)=U(x,y)(e^{x}-e^{y}).$

###  8.2. The factor $A_{3}$ is non-negative

We now show that the first term in the above expression is positive, which is
needed to use our standard bound on the exponent. Since
$A_{3}(x,y)=4(e^{x}-1)V(x,y)$, where

$\displaystyle V(x,y)$
$\displaystyle=4\left(e^{x}x-e^{y}y-e^{x+y}(x-y)\right)^{2}$
$\displaystyle\qquad+3\left(e^{x}-1\right)\left(e^{x}-e^{y}\right)\left(e^{x+y}(3x-3y-2)-e^{x}(3x+2)+2e^{2x}+e^{y}(3y+2)\right),$

thus reduces to the inequality $V(x,y)\geq 0$. We shall expand $V(x,y)$ in $x$
and show that the coefficients are non-negative. The coefficient in front of
$\frac{x^{n}}{n!}$ is of the form

$Q_{n}(y)=\alpha_{0}+e^{y}(\beta_{0}+\beta_{1}y)+e^{2y}(\gamma_{0}+\gamma_{1}y+\gamma_{2}y^{2})$

with

$\alpha_{0}(n)=2^{n}\left(n^{2}+\frac{7n}{2}+6\right)-3^{n+1}(n+4)+6\cdot
4^{n}$
$\beta_{0}(n)=2^{n}\left(-2n^{2}+2n+24\right)-9n+3^{n}(3n-12)-12,\qquad\beta_{1}(n)=-8n-3^{n+2}+2^{n}(4n+18)-9$

and

$\gamma_{0}(n)=2^{n}\left(n^{2}-\frac{11n}{2}+6\right)+9n-12,\qquad\gamma_{1}(n)=8n+2^{n}(9-4n)-18,\qquad\gamma_{2}(n)=2^{n+2}-8,$

whereas $Q_{0}$ and $Q_{1}$ vanish. It is straightforward to check that

$\alpha_{0}(n)>0,\qquad\beta_{1}(n)<0,\qquad\gamma_{0}(n)>0,\qquad\gamma_{2}(n)>0.$

It is also easy to show that $\gamma_{1}(n)<0$ if and only if $n\geq 3$. We
show that $\beta_{0}>0$ if and only if $n\geq 5$. For $n\geq 11$ we have

$\beta_{0}(n)\geq-2\cdot 2^{n}n^{2}+3\cdot 3^{n}$

where the last inequality can be proved by induction. The remaining case
$2\leq n\leq 10$ are proved directly.

Now, for $n\geq 5$ all the terms have correct sign, so $Q_{n}(y)$ in
nonnegative. We also have

$Q_{2}(y)=e^{2y}\left(8y^{2}+2y+2\right)+e^{y}(-2y-4)+2\geq
2e^{2y}(y+1)-2e^{y}(y+2)+2$

It is therefore enough to prove the inequality

$e^{2y}(y+1)-e^{y}(y+2)+1\geq 0.$

If $y<-2$ then the second terms is positive and the first term is already
smaller in absolute value than $1$, so the inequality holds. If $y\in[-2,0]$
then we shall show that the function is decreasing. The derivative is
$e^{y}\left(-y+e^{y}(2y+3)-3\right)$ so it is enough to show that
$-y+e^{y}(2y+3)<3.$ The second derivative of the left hand side is
$e^{y}(2y+7)>0$, so by convexity it is enough to verify the inequality for
$y=-2,0$.

We have

$Q_{3}(y)=e^{2y}\left(24y^{2}-18y+3\right)+e^{y}(-36y-24)+21\geq
3e^{2y}-24e^{y}+21=3(e^{2y}-8e^{y}+7).$

Let $t=e^{y}\in[0,1]$. We want to show $t^{2}-8t+7=(t-1)(t-7)>0$, which is
true. Finally

$Q_{4}(y)=e^{2y}\left(56y^{2}-98y+24\right)+e^{y}(-226y-48)+168>-48e^{y}+168\geq-48+168>0.$

After using the bound on the exponent, we are left with proving

$A_{3}(x,y)+C(x,y)B_{3}(x,y)\geq 0.$

###  8.3. Taylor expansion

We expand

$L(x,y):=A_{3}(x,y)+C(x,y)B_{3}(x,y)$

in $x$ using the command

$R[n\textunderscore,y\textunderscore]:=\textrm{SeriesCoefficient}[\textrm{Expand}[L[x,y]],\\{x,0,n\\}].$

The coefficient in front of $\frac{x^{n}}{n!}$ is of the form

$P_{n}(y)=a_{0}+e^{y}(b_{0}+b_{1}y+b_{2}y^{2})+e^{2y}(c_{0}+c_{1}y+c_{2}y^{2})+e^{3y}(d_{0}+d_{1}y+d_{2}y^{2})+e^{4y}e_{0},$

for $n\geq 2$ (we have $P_{0}=P_{1}=P_{2}\equiv 0$), where

$\displaystyle a_{0}(n)$
$\displaystyle=4^{n}\left(\frac{9n^{2}}{8}-\frac{57n}{8}-36\right)+3^{n}\left(\frac{16n^{2}}{9}+\frac{200n}{9}+60\right)+2^{n}\left(-4n^{2}-14n-24\right)$
$\displaystyle e_{0}(n)$ $\displaystyle=-6n^{2}+18n-12,$

whereas the coefficients $b_{i}$ are given by

$\displaystyle b_{0}(n)$
$\displaystyle=2^{n}\left(\left(11-\frac{e^{2}}{2}\right)n^{2}+\left(\frac{e^{2}}{2}-29\right)n-12e^{2}-108\right)$
$\displaystyle\qquad+4^{n}\left(\left(\frac{15}{8}-\frac{e^{2}}{4}\right)n^{2}+\left(\frac{33}{8}+\frac{e^{2}}{4}\right)n-24e^{2}+36\right)$
$\displaystyle\qquad+3^{n}\left(\left(\frac{e^{2}}{27}-\frac{2}{9}\right)n^{4}+\left(\frac{8}{9}-\frac{4e^{2}}{27}\right)n^{3}+\left(\frac{77e^{2}}{27}-\frac{44}{3}\right)n^{2}+\left(\frac{34e^{2}}{27}-10\right)n+36e^{2}+24\right)$
$\displaystyle\qquad+(36n+48)$ $\displaystyle b_{1}(n)$
$\displaystyle=3^{n}\left(\left(\frac{4}{3}-\frac{2e^{2}}{9}\right)n^{3}+\left(\frac{2e^{2}}{9}-\frac{4}{3}\right)n^{2}+\left(\frac{104}{3}-4e^{2}\right)n+108\right)+2^{n}\left(\left(2e^{2}-44\right)n-108\right)$
$\displaystyle\qquad+4^{n}\left(\left(2e^{2}-12\right)n-36\right)+(32n++36)$
$\displaystyle b_{2}(n)$
$\displaystyle=3^{n}\left(\left(\frac{e^{2}}{3}-2\right)n^{2}+\left(\frac{e^{2}}{3}-2\right)n+6e^{2}-36\right)+\left(12-2e^{2}\right)2^{n}+\left(24-4e^{2}\right)4^{n},$

the coefficients $c_{i}$ equal to

$\displaystyle c_{0}(n)$
$\displaystyle=\left(\left(2e^{2}-12\right)n^{2}+\left(-24-2e^{2}\right)n+12e^{2}+36\right)$
$\displaystyle\qquad+2^{n}\left(\left(\frac{9}{8}-\frac{3e^{2}}{16}\right)n^{4}+\left(\frac{7e^{2}}{8}-\frac{21}{4}\right)n^{3}+\left(\frac{235}{8}-\frac{125e^{2}}{16}\right)n^{2}+\left(\frac{115}{4}+\frac{9e^{2}}{8}\right)n-48e^{2}+72\right)$
$\displaystyle\qquad+3^{n}\left(\left(\frac{e^{2}}{81}-\frac{2}{27}\right)n^{4}+\left(\frac{8}{9}-\frac{4e^{2}}{27}\right)n^{3}+\left(\frac{137e^{2}}{81}-\frac{172}{27}\right)n^{2}+\left(\frac{50}{9}-\frac{50e^{2}}{9}\right)n+36e^{2}-84\right)$
$\displaystyle c_{1}(n)$
$\displaystyle=2^{n}\left(\left(\frac{3e^{2}}{4}-\frac{9}{2}\right)n^{3}+\left(\frac{15}{2}-\frac{5e^{2}}{4}\right)n^{2}+\left(\frac{17e^{2}}{2}-19\right)n-108\right)$
$\displaystyle\qquad+3^{n}\left(\left(\frac{4}{9}-\frac{2e^{2}}{27}\right)n^{3}+\left(\frac{2e^{2}}{3}-4\right)n^{2}+\left(\frac{152}{9}-\frac{124e^{2}}{27}\right)n+36\right)+\left(-8-4e^{2}\right)n+108$
$\displaystyle c_{2}(n)$
$\displaystyle=2^{n}\left(\left(\frac{9}{2}-\frac{3e^{2}}{4}\right)n^{2}+\left(\frac{3}{2}-\frac{e^{2}}{4}\right)n-8e^{2}\right)$
$\displaystyle\qquad+3^{n}\left(\left(\frac{e^{2}}{9}-\frac{2}{3}\right)n^{2}+\left(\frac{14}{3}-\frac{7e^{2}}{9}\right)n+6e^{2}-20\right)+\left(36+2e^{2}\right)$

and finally the coefficients $d_{i}$ equal to

$\displaystyle d_{0}(n)$
$\displaystyle=\left(\left(2e^{2}-30\right)n^{2}+\left(18-2e^{2}\right)n+12e^{2}-72\right)$
$\displaystyle\qquad+2^{n}\left(\left(\frac{3}{8}-\frac{e^{2}}{16}\right)n^{4}+\left(\frac{5e^{2}}{8}-\frac{15}{4}\right)n^{3}+\left(\frac{129}{8}-\frac{55e^{2}}{16}\right)n^{2}+\left(\frac{71e^{2}}{8}-\frac{123}{4}\right)n-12e^{2}+60\right)$
$\displaystyle d_{1}(n)$
$\displaystyle=2^{n}\left(\left(\frac{e^{2}}{4}-\frac{3}{2}\right)n^{3}+\left(\frac{21}{2}-\frac{7e^{2}}{4}\right)n^{2}+\left(\frac{7e^{2}}{2}-21\right)n\right)+\left(24-4e^{2}\right)n$
$\displaystyle d_{2}(n)$
$\displaystyle=2^{n}\left(\left(\frac{3}{2}-\frac{e^{2}}{4}\right)n^{2}+\left(\frac{5e^{2}}{4}-\frac{15}{2}\right)n-2e^{2}+12\right)+\left(2e^{2}-12\right)$

###  8.4. Signs of $c_{0},c_{1},c_{2}$ for $n\geq 4$

Let $n\geq 4$. We show the inequalities

$\qquad\qquad c_{0}(n)>0,\qquad c_{1}(n)<0,\qquad c_{2}(n)>0.$

We start with $c_{0}(n)$. We have

$\displaystyle c_{0}(n)$
$\displaystyle>(2n^{2}-39n+124)+2^{n}\left(-0.27n^{4}+n^{3}-29n^{2}+37n-283\right)$
$\displaystyle\qquad+3^{n}\left(0.01n^{4}-0.21n^{3}+6n^{2}-36n+182\right).$

If we now assume that $n\geq 25$ then we can further estimate

$2n^{2}-39n+124>0,\qquad-0.27n^{4}+n^{3}-29n^{2}+37n-283>-0.27n^{4}-29n^{2}-283>-n^{4}$

and

$6n^{2}-36n+182>0,\qquad 0.01n^{4}-0.21n^{3}>0.001n^{4}$

thus

$c_{0}(n)>-2^{n}n^{4}+3^{n}\cdot 0.001n^{4}=n^{4}(3^{n}\cdot
10^{-3}-2^{n})>0.$

The remaining cases are checked directly.

We now show that $c_{1}(n)<0$. For $n\geq 23$ we have

$\displaystyle c_{1}(n)$
$\displaystyle<3^{n}\left(-0.1n^{3}+n^{2}-17n+36\right)+2^{n}\left(1.05n^{3}-n^{2}+44n-108\right)-37n+108$
$\displaystyle<2^{n}\left(1.05n^{3}+44n\right)+3^{n}\left(-0.1n^{3}+n^{2}+36\right)<2^{n}\cdot
2n^{3}+3^{n}(-0.1n^{3}+2n^{2})$ $\displaystyle<2^{n}\cdot
2n^{3}+3^{n}(-0.1n^{3}+2n^{2})<2^{n}\cdot 2n^{3}-3^{n}\cdot
0.01n^{3}=-n^{3}(3^{n}10^{-3}-2\cdot 2^{n})<0.$

The cases $4\leq n\leq 23$ are checked directly.

Finally, we show that $c_{2}(n)>0$. We can assume $n\geq 9$ since other cases
can be checked directly. We have

$\displaystyle c_{2}(n)$
$\displaystyle>3^{n}\left(0.15n^{2}-1.1n+24\right)-2^{n}\left(1.05n^{2}+0.35n+60\right)>3^{n}\cdot
0.1n^{2}-2^{n}\cdot 2n^{2}$ $\displaystyle=n^{2}(3^{n}\cdot 0.1-2^{n}\cdot
2)>0.$

###  8.5. Bounding terms containing high powers of $e^{y}$

We are going to obtain a useful bound on $P_{n}(y)$ by showing that

$e^{2y}(c_{0}+c_{1}y+c_{2}y^{2})+e^{3y}(d_{0}+d_{1}y+d_{2}y^{2})+e^{4y}e_{0}\geq
0.$

Since $c_{1}(n)<0$ and $c_{2}(n)>0$, it is enough to show that

$c_{0}+e^{y}(d_{0}+d_{1}y+d_{2}y^{2})+e^{2y}e_{0}\geq 0$

for all $n\geq 4$. Since $e^{y}\leq 1$, $|e^{y}y|<1$ and $e^{2y}y^{2}<1$ it is
enough to show that

(10) $c_{0}-(|d_{0}|+|d_{1}|+|d_{2}|+|e_{0}|)>0.$

We have

$|d_{0}|\leq
16n^{2}+4n+17+2^{n}\left(0.1n^{4}+n^{3}+10n^{2}+35n+29\right)<40n^{2}+2^{n}\cdot
2n^{4}<2^{n}\cdot 3n^{4}.$

Moreover

$|d_{1}|<2^{n}\left(0.35n^{3}+2.5n^{2}+5n\right)+6n<2^{n}\cdot
2n^{3}\leq\frac{1}{2}\cdot 2^{n}\cdot n^{4}$

and

$|d_{2}|<2^{n}\left(0.35n^{2}+2n+3\right)+3<2^{n}\cdot 2n^{2}<\frac{1}{8}\cdot
2^{n}n^{4}.$

Also

$|e_{0}|<6n^{2}+18n+12<12n^{2}<\frac{1}{4}2^{n}n^{4}.$

As a result

$|d_{0}|+|d_{1}|+|d_{2}|+|e_{0}|<2^{n}\cdot 4n^{4}$

We have already seen that for $n\geq 25$ one has

$c_{0}\geq n^{4}(3^{n}\cdot 10^{-3}-2^{n}).$

Therefore it is enough to check that $3^{n}\cdot 10^{-3}>5\cdot 2^{n}$, which
is true for $n\geq 25$. For $4\leq n\leq 24$ the inequality (10) is also
valid.

###  8.6. Positivity of $P_{n}(y)$ for $n\geq 8$

According to the previous subsection, it is enough to prove that

$a_{0}+e^{y}(b_{0}+b_{1}y+b_{2}y^{2})\geq 0.$

It turns out that for $n\geq 9$ the inequality follows from the trivial bound

$a_{0}+e^{y}(b_{0}+b_{1}y+b_{2}y^{2})\geq
a_{0}-(|b_{0}|+e^{-1}|b_{1}|+4e^{-2}|b_{2}|).$

The right hand side is positive for $9\leq n\leq 25$. We have

$a_{0}(n)>3^{n}\left(n^{2}+22n+60\right)+4^{n}\left(1.125n^{2}-8n-36\right)-2^{n}\left(4n^{2}+14n+24\right).$

Moreover

$\displaystyle|b_{0}|$ $\displaystyle\leq
4^{n}\left(0.028n^{2}+6n+142\right)+2^{n}\left(8n^{2}+26n+197\right)$
$\displaystyle\qquad+3^{n}\left(0.06n^{4}+0.3n^{3}+7n^{2}+n+291\right)+36n+48$
$\displaystyle|b_{1}|$ $\displaystyle\leq
3^{n}\left(0.4n^{3}+0.4n^{2}+6n+108\right)+32n+4^{n}(3n+36)+2^{n}(30n+108)+36$
$\displaystyle|b_{2}|$ $\displaystyle\leq
3^{n}\left(0.5n^{2}+0.5n+9\right)+3\cdot 2^{n}+6\cdot 4^{n}.$

Putting these crude bounds together gives

$\displaystyle a_{0}$
$\displaystyle-(|b_{0}|+e^{-1}|b_{1}|+4e^{-2}|b_{2}|)\geq
a_{0}-(|b_{0}|+|b_{1}|+|b_{2}|)$ $\displaystyle\qquad\geq
4^{n}\left(1.097n^{2}-17n-220\right)+2^{n}\left(-12n^{2}-70n-332\right)$
$\displaystyle\qquad\qquad+3^{n}\left(-0.06n^{4}-0.7n^{3}-6.9n^{2}+14.5n-348\right)-68n-84.$

If $n\geq 26$ then

$1.097n^{2}-17n-220\geq 0.1n^{2},\qquad 12n^{2}+70n+332<16n^{2},\qquad
68n+84<2^{n}n^{2}$

and

$0.06n^{4}+0.7n^{3}+6.9n^{2}-14.5n+348<0.1n^{4}.$

Using these estimates we get a lower bound

$4^{n}\cdot 0,1n^{2}-2^{n}\cdot 16n^{2}-3^{n}\cdot
0.1n^{4}-2^{n}n^{2}=n^{2}(4^{n}\cdot 0.1-2^{n}\cdot 17-3^{n}\cdot 0.1n^{2})$

Note that

$4^{n}\cdot 0.1-2^{n}\cdot 17-3^{n}\cdot 0.1n^{2}\geq 4^{n}\cdot
0.1-3^{n}\cdot 0.2n^{2}>0.$

It remains to prove the inequality in the case $n=8$, in which case

$1+e^{y}\left(\frac{b_{0}}{a_{0}}+\frac{b_{1}}{a_{0}}y+\frac{b_{2}}{a_{0}}y^{2}\right)>1-e^{y}(1+y+0.12y^{2}).$

The right hand side is clearly positive for, say, $y<-5$. The derivative of
the right hand side is $e^{y}(-0.12y^{2}-1.24y-2.)$ which on the interval
$[-5,0]$ has sign pattern $(+,-)$ and thus it is enough to check the
inequality for $y=-5$ and $y=0$.

We have $P_{0}\equiv P_{1}\equiv P_{2}\equiv 0$. Positivity of $P_{n}(y)$ for
$3\leq n\leq 7$ is proved in the Appendix. Below we provide a graph of these
functions.

Figure 3. The graphs of
$P_{3}(y),\frac{2}{4!}P_{4}(y),\frac{1}{5!}P_{5}(y),\frac{1}{6!}P_{6}(y)$ and
$\frac{1}{7!}P_{7}(y)$.

## Acknowledgements

The authors express their gratitude to Krzysztof Oleszkiewicz for lengthy
discussions, suggestions, and support for this project, without which this
paper would not exist. We thank the Institut Henri Poincaré and Université
Paris Nanterre, where the majority of this work was completed, for their warm
hospitality. We also extend gratitude to Sergey Bobkov, Mokshay Madiman, and
Arnaud Marsiglietti for generously sharing their thoughts, motivations, and
suggestions about this problem.

## References

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##  9\. Appendix

###  9.1. Functions $P_{n}(y)$ for the first derivative

####  9.1.1. Positivity of $P_{1}(y)$

We have

$\displaystyle 6e^{-y}P_{1}(y)$
$\displaystyle=\left[\left(18-2e^{2}\right)y^{2}-24y-12e^{2}\right]$
$\displaystyle\qquad+e^{y}\left[\left(e^{2}-6\right)y^{4}+\left(12-2e^{2}\right)y^{3}+\left(8e^{2}-24\right)y^{2}+\left(48-12e^{2}\right)y+12e^{2}\right]$
$\displaystyle\qquad+e^{2y}\left[6y^{2}-24y\right].$

The first step is to forget the positive terms including $e^{y}y^{4}$ and
$e^{y}y^{3}$, which leads to

$\displaystyle 6e^{-y}P_{1}(y)$
$\displaystyle\geq\left[\left(18-2e^{2}\right)y^{2}-24y-12e^{2}\right]+e^{y}\left[\left(8e^{2}-24\right)y^{2}+\left(48-12e^{2}\right)y+12e^{2}\right]+e^{2y}\left[6y^{2}-24y\right].$

_Case 1: $y<-3$._ In this case the first bracket is positive and thus the
whole function is positive. _Case 2: $y\in[-\frac{2}{5},0]$._ In this case we
use the bound $e^{t}\geq 1+t+\frac{t^{2}}{2}+\frac{t^{3}}{6}$ which gives

$6e^{-y}P_{1}(y)\geq\frac{2}{3}y^{3}\left(\left(6+2e^{2}\right)y^{2}+\left(3e^{2}-36\right)y+6e^{2}-54\right).$

We have

$\left(6+2e^{2}\right)y^{2}+\left(3e^{2}-36\right)y+6e^{2}-54\leq\left(6+2e^{2}\right)\frac{4}{25}+\left(3e^{2}-36\right)\left(-\frac{2}{5}\right)+6e^{2}-54=-0.808033<0.$

_Case 3: $y\in[-3,-\frac{2}{5}]$._ The derivative of the right hand side in
the above estimate is

$4\left(-6e^{y}\left(y^{2}-2\right)+3e^{2y}\left(y^{2}-3y-2\right)-e^{2}y+e^{y+2}(2y+1)y+9y-6\right)$

and therefore its absolute value is bounded by

$4\left(6\left(y^{2}+2\right)+3\left(y^{2}+3|y|+2\right)+e^{2}|y|+e^{2}(2|y|+1)|y|+9|y|+6\right)\leq
1500.$

Therefore it is enough to show that for
$y_{k}=-3+(-\frac{2}{5}+3)\frac{k}{50000}$ for $k=0,\ldots,50000$ the values
of our lower bound are greater than $\frac{1}{10}$, which is the case.

####  9.1.2. Positivity of $P_{2}(y)$

We have

$\displaystyle P_{2}(y)$
$\displaystyle=e^{3y}\left(\frac{y^{2}}{2}-4y+3\right)+e^{y}\left(\left(\frac{15}{2}-\frac{5e^{2}}{6}\right)y^{2}+\left(\frac{2e^{2}}{3}-16\right)y-5e^{2}+1\right)$
$\displaystyle\qquad+e^{2y}\left(\left(\frac{e^{2}}{4}-\frac{3}{2}\right)y^{4}+\left(7-\frac{7e^{2}}{6}\right)y^{3}+\left(3e^{2}-12\right)y^{2}+\left(20-\frac{17e^{2}}{3}\right)y+5e^{2}-5\right)+1.$

We can again neglect positive terms

$e^{2y}\left(\left(\frac{e^{2}}{4}-\frac{3}{2}\right)y^{4}+\left(7-\frac{7e^{2}}{6}\right)y^{3}\right)\qquad\textrm{and}\qquad
e^{y}\left(\frac{15}{2}-\frac{5e^{2}}{6}\right)y^{2}\qquad\textrm{and}\qquad
e^{3y}\frac{y^{2}}{2}$

getting a bound

$P_{2}(y)\geq
e^{2y}\left(\left(3e^{2}-12\right)y^{2}+\left(20-\frac{17e^{2}}{3}\right)y+5e^{2}-5\right)+e^{y}\left(\left(\frac{2e^{2}}{3}-16\right)y-5e^{2}+1\right)+e^{3y}(3-4y)+1.$

_Case 1. $y<-4$_. In this case $\left(\frac{2e^{2}}{3}-16\right)y-5e^{2}+1>0$
and the rest of the terms are also positive.

_Case 2. $y\in[-\frac{1}{10},0]$_. In this case we can assume that the bracket
multiplied by $e^{y}$ is negative, since otherwise there is nothing to prove.
For this term we are going to use the bound $e^{y}\leq 1+y+\frac{y^{2}}{2}$
and for the other exponents the bound $e^{t}\geq
1+t+\frac{t^{2}}{2}+\frac{t^{3}}{6}$, yielding

$P_{2}(y)\geq\frac{y^{2}}{18}\left(\left(72e^{2}-288\right)y^{3}+\left(-276-28e^{2}\right)y^{2}+\left(30e^{2}-57\right)y-3e^{2}+72\right).$

The polynomial

$\left(72e^{2}-288\right)y^{3}+\left(-276-28e^{2}\right)y^{2}+\left(30e^{2}-57\right)y-3e^{2}+72$

clearly increases with $y$ (checking the signs of the coefficients) at thus it
is enough to show that it is positive for $y=\frac{1}{10}$, which is the case.

_Case 3. $y\in[-4,-\frac{1}{10}]$_. The derivative of the estimate is

$-\frac{1}{3}e^{y}\left(6e^{y}\left(12y^{2}-8y-5\right)+e^{y+2}\left(-18y^{2}+16y-13\right)+48y+3e^{2y}(12y-5)+e^{2}(13-2y)+45\right)$

and thus its absolute value can be upper bounded by

$\frac{1}{3}\left(6\left(12y^{2}+8|y|+5\right)+e^{2}\left(18y^{2}+16|y|+13\right)+48|y|+3(12|y|+5)+e^{2}(13+2|y|)+45\right)$

which is maximized for $y=-4$ and does not exceed $1600$. Therefore it is
enough to show that for $y_{k}=-4+(-\frac{1}{10}+4)\frac{k}{10^{6}}$ for
$k=0,\ldots,10^{6}$ the values of our lower bound are greater than $0.02$,
which is the case.

####  9.1.3. Positivity of $P_{3}(y)$

We have

$\displaystyle P_{3}(y)$
$\displaystyle=3+e^{3y}\left(\frac{y^{2}}{6}-2y+3\right)+e^{y}\left(\left(\frac{59}{6}-\frac{10e^{2}}{9}\right)y^{2}+\left(\frac{5e^{2}}{3}-28\right)y-7e^{2}+7\right)$
$\displaystyle\qquad+e^{2y}\left(\left(\frac{7e^{2}}{36}-\frac{7}{6}\right)y^{4}+\left(\frac{26}{3}-\frac{13e^{2}}{9}\right)y^{3}+\left(\frac{37e^{2}}{9}-20\right)y^{2}+\left(28-\frac{23e^{2}}{3}\right)y+7e^{2}-13\right).$

We can neglect all the terms having powers of $y$ greater than $1$. We get

$P_{3}(y)\geq
e^{3y}\left(-2y+3\right)+e^{y}\left(\left(\frac{5e^{2}}{3}-28\right)y-7e^{2}+7\right)+e^{2y}\left(\left(28-\frac{23e^{2}}{3}\right)y+7e^{2}-13\right)+3.$

_Case 1. $y<-3$._ In this case the bracket multiplied by $e^{y}$ is positive
and there is nothing to prove.

_Case 2. $y\in[-3,0]$._ We shall show that the bound is decreasing on
$[-3,0]$. The derivative is

$-\frac{1}{3}e^{y}\left(84y+3e^{2y}(6y-7)-6e^{y}(28y+1)+e^{y+2}(46y-19)+e^{2}(16-5y)+63\right).$

It is therefore enough to show that

$84y+3e^{2y}(6y-7)-6e^{y}(28y+1)+e^{y+2}(46y-19)+e^{2}(16-5y)+63$

is positive. The derivative of this expression is

$12e^{2y}(3y-2)-6e^{y}(28y+29)+e^{y+2}(46y+27)-5e^{2}+84$

which in absolute value is bounded by

$12(3|y|+2)+6(28|y|+29)+e^{2}(46|y|+27)+5e^{2}+84<2200.$

Therefore it is enough to show that for $y_{k}=-3\frac{k}{1500}$ for
$k=0,\ldots,1500$ the values of our lower bound are greater than $0.02$, which
is the case.

####  9.1.4. Positivity of $P_{4}(y)$

We have the bound

$e^{y}(-357y-894)+e^{2y}(786-569y)+108$

and the first term is positive for $y<-2.6$. Assume $y\in[-2.6,0]$. It is
enough to show that this function is decreasing on this interval. Its
derivative is

$-e^{y}\left(357y+e^{y}(1138y-1003)+1251\right)$

and thus it is enough to show that

$357y+e^{y}(1138y-1003)+1251\geq 0$

Dividing by $357$ and estimating gives the goal

$y+e^{y}\left(\frac{7y}{2}-3\right)+\frac{7}{2}\geq 0\qquad y\in[-2.6,0]$

The derivative of this function is $\frac{1}{2}e^{y}(7y+1)+1$ whose absolute
value can be upper bounded by $\frac{7|y|}{2}+\frac{3}{2}<10$. Therefore it is
enough to show that for $y_{k}=-2.6\frac{k}{3000}$ for $k=0,\ldots,3000$ the
values of our lower bound are greater than $0.001$, which is the case.

####  9.1.5. Positivity of $P_{5}(y)$

We have

$P_{5}(y)\geq e^{y}(-1277y-2758)+2525e^{2y}+540$

The first term is positive for $y<-2.2$. Let us therefore assume that
$y\in[-2.2,0]$. The derivative of the estimate is

$e^{y}\left(-1277y+5050e^{y}-4035\right)$

So its absolute value is upper bounded by

$1277|y|+5050+4035\leq 12000$

Therefore it is enough to show that for $y_{k}=-2.2\frac{k}{300}$ for
$k=0,\ldots,300$ the values of our lower bound are greater than $100$, which
is the case.

###  9.2. Functions $P_{n}(y)$ for the second derivative

####  9.2.1. Positivity of $P_{2}(y)$

We have

$\displaystyle P_{2}(y)$
$\displaystyle=\frac{2}{3}e^{y}\left(\left(6-e^{2}\right)y^{2}-18y-6e^{2}-6\right)$
$\displaystyle\quad+\frac{2}{3}e^{2y}\left(\left(6-e^{2}\right)y^{3}+\left(6+e^{2}\right)y^{2}+\left(24-6e^{2}\right)y+6e^{2}-6\right)+4e^{3y}(1-y)+4$

_Case 1: $y<-3$._ Note that

$P_{2}(y)\geq\frac{2}{3}e^{y}\left(\left(6-e^{2}\right)y^{2}-18y-6e^{2}-6\right)+4\geq
4-e^{y}(y^{2}+34).$

To prove that this is positive it is enough to observe that it is positive for
$y=-3$ and then show that the expression is decreasing. Equivalently, we want
to show that $e^{y}(y^{2}+34)$ is increasing. The derivative is
$e^{y}\left(y^{2}+2y+34\right)=e^{y}\left((y+1)^{2}+33\right)>0$

_Case 2: $y\in[-\frac{1}{2},0]$._ It is enough to show that the derivative on
this interval is negative. The derivative is of the form
$-\frac{2}{3}e^{y}R(y)$, where

$\displaystyle R(y)$
$\displaystyle=(\left(e^{2}-6\right)y^{2}+\left(6+2e^{2}\right)y+6e^{2}+24)+e^{2y}(18y-12)$
$\displaystyle\qquad+e^{y}(\left(2e^{2}-12\right)y^{3}+\left(e^{2}-30\right)y^{2}+\left(10e^{2}-60\right)y-6e^{2}-12).$

Using estimates $e^{t}\leq 1+t+\frac{t^{2}}{2}$ for $t<0$ we get the bound

$\displaystyle R(y)$
$\displaystyle\geq\frac{1}{2}\left(2e^{2}-12\right)y^{5}+\frac{1}{2}\left(5e^{2}-54\right)y^{4}+\frac{1}{2}\left(16e^{2}-72\right)y^{3}+\frac{1}{2}\left(18e^{2}-180\right)y^{2}+\frac{1}{2}\left(12e^{2}-144\right)y$
$\displaystyle\geq 2y^{5}-8y^{4}+24y^{3}-24y^{2}-27y.$

The positivity of this expression is equivalent with the positivity of

$-2y^{4}+8y^{3}-24y^{2}+24y+27.$

Here the worst case is $y=-\frac{1}{2}$.

_Case 3: $y\in[-3,-\frac{1}{2}]$._ We crudely bound the derivative

$\displaystyle|P_{2}^{\prime}(y)|$
$\displaystyle\leq|R(y)|\leq\left(6+e^{2}\right)y^{2}+\left(\left(12+2e^{2}\right)|y|^{3}+\left(30+e^{2}\right)y^{2}+\left(60+10e^{2}\right)|y|+6e^{2}+12\right)$
$\displaystyle\qquad\qquad\qquad+\left(6+2e^{2}\right)|y|+(18|y|+12)+6e^{2}+24$
$\displaystyle\leq 27y^{3}+51y^{2}+173y+137<2000.$

Therefore it is enough to show that for
$y_{k}=-3+(-\frac{1}{2}+3)\frac{k}{6000}$ for $k=0,\ldots,6000$ the values of
our lower bound are greater than $1$, which is the case.

####  9.2.2. Positivity of $P_{3}(y)$

We have

$\displaystyle P_{3}(y)$
$\displaystyle=\frac{20}{3}+\frac{2}{9}e^{y}\left(\left(24-4e^{2}\right)y^{2}+\left(3e^{2}-96\right)y-24e^{2}+9\right)$
$\displaystyle\qquad+\frac{1}{9}e^{2y}\left(\left(30-5e^{2}\right)y^{3}+\left(14e^{2}-48\right)y^{2}+\left(96-36e^{2}\right)y+48e^{2}-108\right)+\frac{2}{3}e^{3y}(5-2y).$

After neglecting various positive terms we get a bound

$\displaystyle P_{3}(y)$
$\displaystyle\geq\frac{2}{9}e^{y}\left(\left(24-4e^{2}\right)y^{2}+\left(3e^{2}-96\right)y-24e^{2}+9\right)$
$\displaystyle\qquad+\frac{1}{9}e^{2y}\left(\left(96-36e^{2}\right)y+48e^{2}-108\right)+\frac{10e^{3y}}{3}+\frac{20}{3}.$

_Case 1: $y<-2.1$._ Forgetting about additional positive terms we get

$P_{3}(y)\geq\frac{20}{3}+\frac{2}{9}e^{y}\left(\left(24-4e^{2}\right)y^{2}+(-24e^{2}+9)\right)\geq
6-e^{y}(2y^{2}+38).$

The derivative of $e^{y}(2y^{2}+38)$ is $2e^{y}\left(y^{2}+2y+19\right)$ and
it therefore positive. Therefore the bound is decreasing in $y$ and so it is
enough to observe that it is positive for $y=-2.1$.

_Case 2: $y\in[-2.1,0]$._ The derivative of the estimate is
$\frac{2}{9}e^{y}R(y)$ where

$R(y)=(\left(24-4e^{2}\right)y^{2}+\left(-48-5e^{2}\right)y-21e^{2}-87)+45e^{2y}+e^{y}\left(-36e^{2}y+96y+30e^{2}-60\right).$

It is enough to show that $R(y)<0$. The derivative of $R$ is

$R^{\prime}(y)=\left(48-8e^{2}\right)y-5e^{2}-48+90e^{2y}+e^{y}(\left(96-36e^{2}\right)y-6e^{2}+36).$

Thus

$|R^{\prime}(y)|\leq 6+12|y|+90+9+171|y|=105+183|y|<500.$

Therefore it is enough to show that for $y_{k}=-2.1\frac{k}{200}$ for
$k=0,\ldots,200$ the values of $R(y)$ are smaller than $-20$, which is the
case.

####  9.2.3. Positivity of $P_{4}(y)$

We use the bound

$P_{4}(y)\geq a_{0}+e^{y}(b_{0}+b_{1}y+b_{2}y^{2})+e^{2y}c_{0}.$

Therefore, bounding the coefficients crudely

(11) $a_{0}^{-1}P_{5}(y)\geq
e^{y}\left(-0.2y^{2}-1.5y-3.7\right)+2.8e^{2y}+1.$

Assume that $y<-3$. Then

$e^{y}\left(-0.2y^{2}-1.5y-3.7\right)+2.8e^{2y}+1\geq
e^{y}\left(-0.2y^{2}-3.7\right)+1.$

Note that for $y<-3$ one has $e^{y}y^{2}<\frac{1}{2}$ and thus

$|e^{y}\left(-0.2y^{2}-3.7\right)|\leq 0.1+3.7e^{-3}<1.$

and therefore the inequality holds in this case.

For $y\in[-3,0]$ we use the netting argument. The derivative of the right hand
side of (11) is

$e^{y}\left(-0.2y^{2}-1.5y-3.7\right)+e^{y}(-0.4y-1.5)+5.6e^{2y}$

and is therefore upper bounded in absolute value by

$\left(0.2y^{2}+1.5|y|+3.7\right)+(0.4|y|+1.5)+5.6\leq 20.$

Therefore it is enough to show that for $y_{k}=-3\frac{k}{2000}$ for
$k=0,\ldots,2000$ the values of of the right hand side of (11) is greater than
$0.05$, which is the case.

####  9.2.4. Positivity of $P_{5}(y)$

We are going to use the bound

$P_{5}(y)\geq a_{0}+e^{y}(b_{0}+b_{1}y+b_{2}y^{2})+e^{2y}c_{0}.$

Therefore, again bounding the coefficients crudely

$a_{0}^{-1}P_{5}(y)\geq e^{y}\left(-y^{2}-y-3\right)+2e^{2y}+1.$

The right hand side vanishes for $y=0$ and therefore it is enough to show that
the right hand side is decreasing. The derivative is equal to
$e^{y}\left(4e^{y}-4-3y-y^{2}\right)$. To show that $4e^{y}-4-3y-y^{2}$ it is
enough to show that it is increasing. The derivative is $4e^{y}-3-2y$. The
minimum of this function is achieved for $y=\log 2$ at is equal to $-1+2\log
2>0$.

####  9.2.5. Positivity of $P_{6}(y)$

We have

$P_{6}(y)\geq a_{0}+e^{y}(b_{0}+b_{2}y^{2})+e^{2y}c_{0}.$

Therefore, again bounding the coefficients crudely gives

$a_{0}^{-1}P_{6}(y)\geq e^{y}\left(-y^{2}-2\right)+e^{2y}+1.$

It is enough to show that the derivative of the right hand side is negative.
This derivative is equal to $2e^{y}\left(e^{y}-1-y-\frac{y^{2}}{2}\right)$ and
the inequality $e^{y}\leq 1+y+\frac{y^{2}}{2}$ is well known.

###  9.3. Functions $P_{n}(y)$ for the third derivative

Note that for $n=5,6,7$ we have $c_{0}>0$, $c_{1}<0$, $c_{2}>0$ and $e_{0}<0$.
Thus we have the bound

$\displaystyle P_{n}(y)$
$\displaystyle>a_{0}+e^{y}(b_{0}+b_{1}y+b_{2}y^{2})+e^{3y}(c_{0}+c_{1}y+c_{2}y^{2})+e^{3y}(d_{0}+d_{1}y+d_{2}y^{2})+e^{3y}e_{0}$
$\displaystyle=a_{0}+e^{y}(b_{0}+b_{1}y+b_{2}y^{2})+e^{3y}(c_{0}+d_{0}+e_{0}+(c_{1}+d_{1})y+(c_{2}+d_{2})y^{2}).$

It is also true that $c_{1}+d_{1}<0$ and $c_{2}+d_{2}>0$ and thus

$P_{n}(y)\geq a_{0}+e^{y}(b_{0}+b_{1}y+b_{2}y^{2})+e^{3y}(c_{0}+d_{0}+e_{0}).$

####  9.3.1. Positivity of $P_{7}(y)$

We have

$a_{0}^{-1}P_{7}(y)>1+e^{y}\left(-0.15y^{2}-1.6y-1.6\right)+1.8e^{3y}>0.9-1.6e^{y}+1.8e^{3y}$

It is therefore enough to show that for $t\in[0,1]$ one has
$0.9-1.6t+1.8t^{3}>0$. By AM-GM we have

$0.9+1.8t^{3}=0.45+0.45+1.8t^{3}>3t\sqrt[3]{0.45^{2}\cdot 1.8}>1.6t$

####  9.3.2. Positivity of $P_{6}(y)$

We have

$a_{0}^{-1}P_{6}(y)>1+e^{y}\left(-0.2y^{2}-2y-2.8\right)+2.8e^{3y}>0.89+e^{y}\left(-2y-2.8\right)+2.8e^{3y}$

The right hand side is clearly positive for $y<-1.4$. Assume $y\in[-1.4,0]$.
Then the derivative of the right hand side is
$e^{y}\left(-2y+8.4e^{2y}-4.8\right)$ and can be upper bounded in absolute
value by $20$. Therefore it is enough to show that for
$y_{k}=-1.4\frac{k}{200}$ for $k=0,\ldots,200$ the values of the right hand
side are greater than $0.2$, which is the case.

####  9.3.3. Positivity of $P_{5}(y)$

We have

(12) $a_{0}^{-1}P_{5}(y)>1+e^{y}\left(-0.24y^{2}-3.45y-4.68\right)+4.3e^{3y}$

We first assume that $y<-2$. In this case

$1+e^{y}\left(-0.24y^{2}-3.45y-4.68\right)+4.3e^{3y}>(1-0.24e^{y}y^{2})+e^{y}(-3.45y-4.68)>0$

Assume $y\in[-2,0]$. The derivative of the right hands side of (12) is

$e^{y}\left(-0.24y^{2}-3.93y+12.9e^{2y}-8.13\right)$

and its absolute value can be upper bounded by $25$. Therefore it is enough to
show that for $y_{k}=-2\frac{k}{5000}$ for $k=0,\ldots,5000$ the values of the
right hand side are greater than $0.01$, which is the case.

####  9.3.4. Positivity of $P_{4}(y)$

Let us first assume that $y<-2.5$. Then

$P_{4}(y)>1+e^{3y}\left(-0.13y^{2}+0.15y-2\right)+e^{y}\left(-0.27y^{2}-7\right)-0.24e^{4y}$

and

$\displaystyle|e^{3y}\left(-0.13y^{2}+0.15y-2\right)+e^{y}\left(-0.27y^{2}-7\right)-0.24e^{4y}|$
$\displaystyle\qquad\leq 0.005\cdot 0.13+0.002\cdot 0.15+0.002+0.52\cdot
0.27+0.58+0.24\cdot 0.001<0.73<1$

Therefore the inequality holds in this case. Let $y\in[-1,0]$. We have the
bound $P_{4}(y)\geq g(y)$ where

$g(y)=a_{0}+e^{y}(b_{0}+b_{1}y+b_{2}y^{2})+e^{2y}c_{0}+e^{3y}(d_{0}+d_{1}y+d_{2}y^{2})+e^{4y}e_{0}$

We are going to show that $g^{\prime}(y)<0$ on $[-1,0]$. We have

$e^{-y}g^{\prime}(y)=(b_{0}+b_{1}+(b_{1}+2b_{2})y+b_{2}y^{2})+e^{y}2c_{0}+e^{2y}(3d_{0}+d_{1}+(3d_{1}+2d_{2})y+3d_{2}y^{2})+e^{3y}4e_{0}$

and

$(e^{-y}g^{\prime}(y))^{\prime}=b_{1}+2b_{2}+2b_{2}y+e^{y}2c_{0}+e^{2y}(6d_{0}+5d_{1}+2d_{2}+(6d_{1}+10d_{2})y+6d_{2}y^{2})+e^{3y}12e_{0}$

We therefore have

$|(e^{-y}g^{\prime}(y))^{\prime}|\leq|b_{1}|+4|b_{2}|+2|c_{0}|+6|d_{0}|+11|d_{1}|+18|d_{2}|+12|e_{0}|<6013$

Therefore it is enough to show that for $y_{k}=-\frac{k}{100}$ for
$k=0,\ldots,100$ the values of $e^{-y}g^{\prime}(y)$ are smaller than $-100$,
which is the case.

We are now left with $y\in[-2.5,-1]$. We shall show that $g(y)>0$. We have

$|g^{\prime}(y)|\leq|b_{0}|+2|b_{1}|+3|b_{2}|+2|c_{0}|+3|d_{0}|+4|d_{1}|+5|d_{2}|+4|e_{0}|<6113$

Therefore it is enough to show that for $y_{k}=-2.5+(-1+2.5)\frac{k}{200}$ for
$k=0,\ldots,200$ the values of $e^{-y}g^{\prime}(y)$ are greater than $100$,
which is the case.

####  9.3.5. Positivity of $P_{3}(y)$

Let us first assume that $y\in[-0.1,0]$. Then it is easy to check that

$b_{0}+b_{1}y+b_{2}y^{2}<0,\qquad c_{0}+c_{1}y+c_{2}y^{2}>0,\qquad
d_{0}+d_{1}y+d_{2}y^{2}<0,\qquad e_{0}<0$

and therefore using standard bounds $e^{t}\leq 1+t+\frac{1}{2}t^{2}$ and
$e^{t}\geq 1+t+\frac{1}{2}t^{2}+\frac{1}{6}t^{3}$ for $t<0$ we get

$\displaystyle P_{3}(y)$
$\displaystyle\geq\frac{4}{3}\left(72+4e^{2}\right)y^{5}+\left(2\left(72+4e^{2}\right)+5\left(12-2e^{2}\right)+32\right)y^{4}$
$\displaystyle\qquad+\left(2\left(72+4e^{2}\right)+32e^{2}+4\left(12-2e^{2}\right)-28\right)y^{3}+\left(2\left(12-2e^{2}\right)-8e^{2}+120\right)y^{2}$
$\displaystyle\geq y^{2}(136y^{3}+221y^{2}+401y+55)>y^{2}(0.136-40.1+55)>0.$

Assume now that $y<-2.5$. We have

$P_{3}(y)\geq
e^{y}\left(-3y^{2}-89\right)+e^{3y}\left(-3y^{2}-41\right)+24e^{2y}y-12e^{4y}+12$

and

$e^{y}\left(3y^{2}+89\right)+e^{3y}\left(3y^{2}+41\right)+24e^{2y}|y|+12e^{4y}$

This expression is increasing in $y$ for $y<-2.5$ and thus it is maximized for
$y=-2.5$ in which case the value is smaller than $10$.

We are now left with the case $y\in[-2.5,-0.1]$. We have

$\displaystyle P_{3}^{\prime}(y)$
$\displaystyle=e^{2y}\left(\left(144+8e^{2}\right)y^{2}+\left(192+8e^{2}\right)y+48e^{2}-72\right)+e^{y}\left(\left(12-2e^{2}\right)y^{2}-4e^{2}y-12e^{2}-24\right)$
$\displaystyle\qquad+e^{3y}\left(\left(36-6e^{2}\right)y^{2}+\left(24-4e^{2}\right)y-36e^{2}+144\right)-48e^{4y}$

Therefore

$|P_{3}^{\prime}(y)|\leq
144+8e^{2}+192+8e^{2}+48e^{2}-72+|12-2e^{2}|+4e^{2}+12e^{2}+24+|36-6e^{2}|+|24-4e^{2}|+|-36e^{2}+144|+48$

This is smaller than $1070$. Therefore it is enough to show that for
$y_{k}=-2.5+(-0.1+2.5)\frac{k}{7000}$ for $k=0,\ldots,7000$ the values of
$e^{-y}g^{\prime}(y)$ are greater than $0.4$, which is the case.