<s>[INST] Prove that $\frac{a}{\sqrt{a^2+b^2}}+\frac{b}{\sqrt{b^2+c^2}}+\frac{c}{\sqrt{c^2+d^2}}+\frac{d}{\sqrt{d^2+a^2}}\leq3$ Prove that if $a,b,c,d$ are positive reals we have:
$$\frac{a}{\sqrt{a^2+b^2}}+\frac{b}{\sqrt{b^2+c^2}}+\frac{c}{\sqrt{c^2+d^2}}+\frac{d}{\sqrt{d^2+a^2}}\leq3.$$
I think that I have found a equality case, for example, when $a=x^4,b=x^3,c=x^2,d=x$, and as $x$ tends to $\infty$, the LHS tends to $3$, but this means that the inequality is very unlikely to be solved with traditional methods, such as Cauchy-Schwartz (my starting idea), so I got stuck.
 [/INST] We first prove the following lemma.
Lemma. Let $\lambda>0$. Then
$$\sup_{x>0}\left(\frac1{\sqrt{1+x}}+\frac1{\sqrt{1+\frac{\lambda}x}}\right)=\begin{cases}\frac2{\sqrt{1+\sqrt\lambda}}&\text{if }\lambda<4\\\sqrt{\frac{\lambda}{\lambda-1}}&\text{if }\lambda\geq 4.\end{cases}$$
Proof. Fixing $\lambda$ constant, let the left side be $f(x)$. The function $f$ is clearly continuous and infinitely differentiable, and so its global extrema are reached when $x\to 0$, $x\to\infty$, or when $f''(x)=0$. It is not hard to compute that
$$f''(x)=0\Longleftrightarrow \lambda^2(1+x)^3=x(\lambda+x)^3;$$
the polynomial on the right factors as $(x^2-\lambda)(x^2-(\lambda^2-3\lambda)x+\lambda)$. So, the extrema are when $x=\sqrt{\lambda}$, or when $x$ and $\lambda/x$ satisfy $x+\frac\lambda x=\lambda^2-3\lambda$. For $\lambda<4$, $x+\frac\lambda x$ is always larger than $\lambda^2-3\lambda$, so the only $x$ with $f''(x)=0$ is $x=\sqrt\lambda$, giving the desired bound. For $\lambda\geq 4$, such a real $x>0$ does exist, and satisfies, letting $y=\lambda/x$ so that $x+y=\lambda^2-3\lambda$,
\begin{align*}
\left(\frac1{\sqrt{1+x}}\right)\left(\frac1{\sqrt{1+y}}\right)&=\frac{1}{\sqrt{1+(\lambda^2-3\lambda)+\lambda}}=\frac1{\lambda-1}\\
\left(\frac1{\sqrt{1+x}}\right)^2+\left(\frac1{\sqrt{1+y}}\right)^2&=\frac{2+x+y}{(1+x)(1+y)}=\frac{\lambda-2}{\lambda-1},
\end{align*}
so
$$\frac1{\sqrt{1+x}}+\frac1{\sqrt{1+y}}=\sqrt{\frac{\lambda-2}{\lambda-1}+\frac2{\lambda-1}}=\frac{\lambda}{\lambda-1}.$$
For $\lambda\geq 4$, this is always at least the value $\frac2{\sqrt{1+\sqrt\lambda}}$ achieved at $x=\sqrt\lambda$, and so is the maximum of $f$. $\square$

Now, under the substitution $w=\frac{b^2}{a^2}$, $x=\frac{c^2}{b^2}$, $y=\frac{d^2}{c^2}$, $z=\frac{a^2}{d^2}$, the problem becomes to show
$$\frac1{\sqrt{1+w}}+\frac1{\sqrt{1+x}}+\frac1{\sqrt{1+y}}+\frac1{\sqrt{1+z}}\leq 3$$
for $w,x,y,z>0$ and $wxyz=1$. If $wx,yz\leq 4$, then the sum is at most
$$\frac2{\sqrt{1+t}}+\frac2{\sqrt{1+\frac1t}}\leq \frac4{\sqrt{2}}=2\sqrt2<3$$
for $t=\sqrt{wx}$, where we have used our lemma on $\lambda=t^2=wx$, $\lambda=1/t^2=yz$, and again on $\lambda=1$. Otherwise, assume without loss of generality that $wx\geq 4$, and let $s=\sqrt{wx}$. Then, by our lemma, the sum is at most
$$\sqrt{\frac{s^2}{s^2-1}}+\frac{2}{\sqrt{1+\frac1s}},$$
where $s\geq 2$. However, for $s\geq 2$,
\begin{align*}
\sqrt{\frac{s^2}{s^2-1}}+\frac{2}{\sqrt{1+\frac1s}}
&=\frac{\sqrt s(\sqrt s+2\sqrt{s-1})}{\sqrt{s^2-1}}\\
&\leq \frac{\sqrt s\left(\sqrt s+2\left[\sqrt s-\frac1{2\sqrt s}\right]\right)}{\sqrt{s^2-1}}\\
&=\frac{3s-1}{\sqrt{s^2-1}},
\end{align*}
which is at most $3$ whenever $s\geq 5/3$ since
$$(3s-1)^2=9s^2-6s+1\leq 9s^2-9\implies \frac{3s-1}{\sqrt{s^2-1}}\leq 3.$$
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