<s>[INST] Prove that $x_1^2+x_2^2+x_3^2=1$ yields $ \sum_{i=1}^{3}\frac{x_i}{1+x_i^2} \le \frac{3\sqrt{3}}{4} $ Prove this inequality, if $x_1^2+x_2^2+x_3^2=1$:
$$ \sum_{i=1}^{3}\frac{x_i}{1+x_i^2} \le \frac{3\sqrt{3}}{4} $$
So far I got to $x_1^4+x_2^4+x_3^4\ge\frac{1}3$ by using QM-AM for $(2x_1^2+x_2^2, 2x_2^2+x_3^2, 2x_3^2+x_1^2)$, but to be honest I'm not sure if that's helpful at all.
(You might want to "rename" those to $x,y,z$ to make writing easier):
$$ \frac{x}{1+x^2}+\frac{y}{1+y^2}+\frac{z}{1+z^2} \le \frac{3\sqrt3}4$$
 [/INST] We assume that $x_i\geq 0$ and let $\theta_i\in \left(0,\frac{\pi}2\right)$ sucht that $x_i=\tan\frac{\theta_i}2$. We have $\sin(\theta_i)=\frac{2x_i}{1+x_i^2}$ and since $\sin$ in concave on $\left(0,\frac{\pi}2\right)$, we have 
$$\sum_{i=1}^3\frac{x_i}{1+x_i^2}=\frac 32\sum_{i=1}^3\frac 13\sin(\theta_i)\leq \frac 32\sin\frac{\theta_1+\theta_2+\theta_3}3.$$
We have, using the convextiy of $x\mapsto \tan^2 x$:
$$\frac 13=\frac 13\sum_{i=1}^3\tan^2\frac{\theta_i}2\geq \tan^2\frac{\theta_1+\theta_2+\theta_3}6,$$ 
so $\tan\frac{\theta_1+\theta_2+\theta_3}6\leq \frac 1{\sqrt 3}$ and $\frac{\theta_1+\theta_2+\theta_3}6\leq \frac{\pi}6$. Finally 
$$\sum_{i=1}^3\frac{x_i}{1+x_i^2}\leq \frac 32\sin \frac{\pi}3=\frac{3\sqrt 3}4,$$
with equality if and only if $(x_1,x_2,x_3)=\left(\frac 1{\sqrt 3},\frac 1{\sqrt 3},\frac 1{\sqrt 3}\right)$.
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