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.

Again partial hint :
Using the formula :
$$\cos\left(\operatorname{arccot}\left(\frac{a}{b}\right)\right)-\sqrt{\frac{a^{2}}{a^{2}+b^{2}}}=0$$
We use Karamata's inequality with $-\cos(x)$:
$$\frac{a}{b}\leq \frac{b}{c}\leq \frac{c}{d}\leq \frac{d}{a}$$
And $$3=4\cos\left(\operatorname{arccot}\left(\frac{3}{\sqrt{7}}\right)\right)$$
With constraint :
$$f(a,b)\geq \operatorname{arccot}\left(\frac{3}{\sqrt{7}}\right),\quad,f(a,b)+f(b,c)\geq 2 \operatorname{arccot}\left(\frac{3}{\sqrt{7}}\right),f(a,b)+f(b,c)+f(c,d)\geq 3\operatorname{arccot}\left(\frac{3}{\sqrt{7}}\right),\quad f\left(a,b\right)+f\left(b,c\right)+f\left(c,d\right)+f\left(d,a\right)\geq4\operatorname{arccot}\left(\frac{3}{\sqrt{7}}\right)$$
Where :
$$f\left(x,y\right)=\operatorname{arccot}\left(\frac{x}{y}\right)$$
