<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] Here's a solution with tangent line method.
Note that $$\frac{x}{1+x^2} \le \frac{3\sqrt{3}}{16}(x^2+1) \iff \frac{1}{48}(3x-\sqrt{3})^2(\sqrt{3}x^2+2x+3\sqrt{3}) \ge 0 \text{ (true by discriminant) }$$
Now we have $$\frac{x}{1+x^2}+\frac{y}{1+y^2}+\frac{z}{1+z^2} \le \frac{3\sqrt{3}}{16}(x^2+y^2+z^2+3) = \frac{3\sqrt{3}}{4}$$ as desired. $\blacksquare$
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