<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] Another solution.  By the AM-GM inequality, we have
$$
\frac{ 2 \cdot \dfrac{1}{\sqrt{3}} \cdot  x_i } {1+x_i^2}
\le
\frac{\dfrac{1}{3} + x_i^2 }{1+x_i^2} = 1 - \frac{2/3}{1+x_i^2},
$$
But $-1/(1+z)$ is a concave function of $z = x^2$, so
\begin{align}
\frac{ \dfrac{2}{\sqrt{3}} x_i } {1+x_i^2}
&\le
\sum_{i=1}^3 \left[1 - \frac{2/3}{1+x_i^2}\right] \\
&\le 3 \left[1 - \frac{2/3}{1 + \dfrac{1}{3}\left(\sum_{i=1}^3 x_i^2\right) } \right] = \frac{3}{2}.
\end{align}
Multiplying both sides by $\sqrt3/2$, we get the desired result.
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