<s>[INST] Prove the inequality $\frac{a}{1+a^2}+\frac{b}{1+b^2}+\frac{c}{1+c^2}\le\frac{3\sqrt3}{4}$ 
Let $a,b,c$ are nonnegative real numbers such that $a^2+b^2+c^2=1$.
Prove the inequality
$$\frac{a}{1+a^2}+\frac{b}{1+b^2}+\frac{c}{1+c^2}\le\frac{3\sqrt3}{4}$$

I tried the method of Lagrange multipliers and Jensen's inequality but I have not been proved this inequality
 [/INST] Let $a=\frac{x}{\sqrt3}$, $b=\frac{y}{\sqrt3}$ and $c=\frac{z}{\sqrt3}.$
Thus, $x^2+y^2+z^2=3$ and we need to prove that
$$\sum_{cyc}\frac{x}{3+x^2}\leq\frac{3}{4}$$ or
$$\sum_{cyc}\left(\frac{1}{4}-\frac{x}{3+x^2}\right)\geq0$$ or
$$\sum_{cyc}\frac{(x-1)(x-3)}{3+x^2}\geq0$$ or
$$\sum_{cyc}\left(\frac{(x-1)(x-3)}{3+x^2}+\frac{x^2-1}{4}\right)\geq0$$ or
$$\sum_{cyc}\frac{(x-1)^2(x^2+2x+9)}{3+x^2}\geq0,$$
which is obvious.
We see that the original inequality is true for all reals $a$, $b$ and $c$ such that $a^2+b^2+c^2=1.$
</s>