<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] Since the inequality is both homogeneous and cyclic, assume that $a = \max(a, b, c, d) = 1$.
It suffices to prove that, for all $b, c, d \in (0, 1]$,
$$\frac{1}{\sqrt{1+b^2}}+\frac{b}{\sqrt{b^2+c^2}}+\frac{c}{\sqrt{c^2+d^2}}+\frac{d}{\sqrt{d^2+1}} \le 3.$$
To this end, first, we have
\begin{align*}
 \frac{1}{\sqrt{1+b^2}}+\frac{b}{\sqrt{b^2+c^2}}
 &= \sqrt{\frac{1}{1+b^2} + \frac{b^2}{b^2+c^2} + \frac{2b}{\sqrt{(1 + b^2)(b^2 + c^2)}}}\\
 &\le \sqrt{\frac{1}{1+b^2} + \frac{b^2}{b^2+c^2} + \frac{2b}{b + bc}}\\
 &= \sqrt{\frac{1}{1+b^2} + \frac{b^2}{b^2+c^2} + \frac{2}{1 + c}}\\
 &\le 1 + \frac14 \left(\frac{1}{1+b^2} + \frac{b^2}{b^2+c^2} + \frac{2}{1 + c}\right)\\
 &\le 1 + \frac14\left(\frac{2}{1 + c} + \frac{2}{1 + c}\right)\\
 &= 1 + \frac{1}{1 + c} \tag{1}
\end{align*}
where we have used Cauchy-Bunyakovsky-Schwarz inequality,
and $\sqrt{u} \le 1 + \frac{u}{4}$,
and $\frac{2}{1+c} - \frac{1}{1+b^2} - \frac{b^2}{b^2+c^2} = \frac{(1-c)(b^2-c)^2}{(1+c)(b^2+1)(b^2+c^2)}\ge 0$.
Second, we have
\begin{align*}
 \frac{c}{\sqrt{c^2+d^2}}+\frac{d}{\sqrt{d^2+1}} &= \sqrt{\frac{c^2}{c^2+d^2}
  + \frac{d^2}{d^2+1} + \frac{2cd}{\sqrt{(c^2+d^2)(d^2+1)}}}\\
 &\le \sqrt{\frac{c^2}{c^2+d^2}
  + \frac{d^2}{d^2+1} + \frac{2cd}{cd + d}}\\
 &= \sqrt{\frac{c^2}{c^2+d^2}
  + \frac{d^2}{d^2+1} + \frac{2c}{c + 1}}\\
 &\le \frac12 + \frac12 \left( \frac{c^2}{c^2+d^2}
 + \frac{d^2}{d^2+1} + \frac{2c}{c + 1}\right)\\
 &\le \frac12 + \frac12 \left( \frac{1}{1+d^2}
 + \frac{d^2}{d^2+1} + \frac{2c}{c + 1}\right)\\
 &= \frac12 + \frac12 \left( 1 + \frac{2c}{c + 1}\right)\\
 &= 1 + \frac{c}{c + 1} \tag{2}
\end{align*}
where we have used Cauchy-Bunyakovsky-Schwarz inequality,
and $\sqrt{u} \le \frac12 + \frac12 u$.
Using (1) and (2), the desired result follows.
We are done.
</s>