<s>[INST] Deducing an inequality in sharpening results in an analytic number theory paper I am studying a research paper in analytic number theory 
It involves an inequality which is to be used in sharpening a result. Unfortunately I am unable to see how to get the required inequality (See proposition $1$) Or read below

$a$ is odd integer $\geq$3 and $r$ is an integer lying between $1\le r<a/2$ . Author deduced $(12)$


French to English in 2nd line after $(12)$ is 

and $s_0$ is the only root in $(0, 1)$ of polynomial $Q(s) =~\dots$

In particular $\delta(a) \geq~\dots$ 

Now, inequality to be deduced is $(13)$ of image using $$\phi_{r, a}\geq \frac{2^{r+1}} { r^{a-2r}}\text{ and }2r\leq{2r+1}\leq{2(r+1) }.$$ 

Taking $\log$ both sides in $\phi_{r, a} $ and putting it in $(12)$ I got RHS of $(12)$ equals 
$$\frac{ (2a-3r-1) \log2 + (2r+1) \log(2r+1) +(a-2r) \log r } { a + (a-2r) \log2+(2r+1)\log(2r+1) } $$ 
which is not equal to $(13)$. 
Can someone please tell how to derive it to $(13)$?
 [/INST] Consider the numerator and denominator of RHS in (12). We have
\begin{align}
&(a-2r)\log(2) + (2r+1)\log(2r+1) - \log(\varphi_{r,a})\\
\ge\ & (a-r)\log(2) - r\log(2) + (2r+1)\log (2r) - \log(\varphi_{r,a})\\
=\ & (a-r)\log(2) + (r+1)\log(2) + (2r+1)\log(r) - \log(\varphi_{r,a})
\end{align}
and
\begin{align}
&a + (a-2r)\log(2) + (2r+1)\log(2r+1)\\
\le\ & a + 1 + (a-2r)\log(2) + (2r+1)\log (2r+2)\\
=\ & (a+1) + (a+1)\log(2) + (2r+1)\log(r+1).
\end{align}
Thus, to prove (13), it suffices to prove that
$$(r+1)\log(2) + (2r+1)\log(r) - \log(\varphi_{r,a}) \ge (a+1)\log(r)$$
or
$$\frac{2^{r+1}}{r^{a-2r}} \ge \varphi_{r,a}.$$
Consider $Q(s) = rs^{a+2} - (r+1)s^{a+1} + (r+1)s - r$.
Note that $Q(0) < 0$, $Q(1)=0$, and
\begin{align}
Q(1 - \tfrac{1}{r}) = -2 (1-\tfrac{1}{r})^{a+1} - \tfrac{1}{r} < 0.
\end{align}
Thus, $s_0 > 1 - \frac{1}{r}$. 
Now we are ready to prove that $\frac{2^{r+1}}{r^{a-2r}} \ge \varphi_{r,a}$.
We have to prove that
$$2^{r+1} \ge ((r+1)s_0 - r)^r(r+1-rs_0)^{r+1} (r - rs_0)^{a-2r}. \tag{1}$$
Since $(r+1)s_0 - r < r+1 - r = 1$ and $r+1 - rs_0 \le r + 1 - r(1-\tfrac{1}{r}) = 2$
and $r - rs_0 \le r - r(1 - \tfrac{1}{r}) = 1$, (1) is true. 
We are done.
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