<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 :
For $x,y>0$ and $x,y\leq 1$ such that $\left(y\right)^{2}+\left(x-\frac{1}{5}\right)^{2}\geq\frac{2}{50}$ we have the inequality :
$$\sqrt{\frac{x^{2}}{y^{2}+x^{2}}}-\left(f\left(x\right)-f\left(y\right)\right)-\frac{3}{4}\le0$$
Where :
$$f\left(x\right)=x\sqrt{\frac{1}{x^{2}+1}}$$
Summing gives the inequality
Edit : we supposed that for :
$$h\left(x,y\right)=\frac{3x^{2}}{2\left(x^{2}+y^{2}\right)} $$
We have :
$$h\left(a,b\right)+h\left(b,c\right)+h\left(c,d\right)+h\left(d,a\right)\leq3$$
Wich can be done since the inequality is homogeneous .
Now we have the inequality for $x,y>6$ and $x/5\leq y$  :
$$\sqrt{\frac{x^{2}}{x^{2}+y^{2}}}-f\left(x\right)+f\left(y\right)-\frac{3x^{2}}{2\left(x^{2}+y^{2}\right)}\le0$$
Where :
$$f\left(x\right)=0.5\ln\left(\sqrt{\frac{x}{x^{2}+1}}\right)$$
I haven't a proof yet but I come back again on it .
Hope it helps .
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