<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] Partial hint :
We have a two first  not hard to show :
Let $x>0$ then define :
$$f\left(x\right)=\frac{x}{\sqrt{x^{2}+1}}$$
Then :
$$f''(x)\leq 0$$
Second fact :
Let $x>0$ then define :
$$(f(e^x))''<0$$
So we use Jensen's inequality with the constraint $c\geq d\ge a\geq 1$ we need  to show for $x,y\leq 1$ :
$$2f\left(\frac{\left(x+y\right)}{2}\right)+2f\left(\sqrt{z}\right)\le3$$
Where $x=\frac{a}{b},y=\frac{b}{c},z=\frac{c}{a}$ and $xyz=1$
Wich is easier and true.


We have the easy inequalities firstly for $0<x\leq y$ :
$$\frac{\sqrt{2}xy}{x^{2}+y^{2}}-\frac{x}{\sqrt{x^{2}+y^{2}}}\geq 0$$
And for $0<y\leq x$ :
$$\frac{\sqrt{2}x^{2}}{x^{2}+y^{2}}-\frac{x}{\sqrt{x^{2}+y^{2}}}\geq 0$$
So assuming $y\geq z\geq  x> 0$ and $d\geq 0$ we have the inequality :
$$g\left(x,y\right)+g\left(y,z\right)+h\left(z,d\right)+h\left(d,x\right)\leq 3$$
$$g\left(x,y\right)=\frac{\sqrt{2}xy}{x^{2}+y^{2}}\quad\quad,h\left(x,y\right)=\frac{x}{\sqrt{x^{2}+y^{2}}}$$
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