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$