<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] Consider the function $f(x)= \frac{\sqrt x}{1+x}$,
with the substitution that $x=a^2,y=b^2,z=c^2$, $\implies x+y+z=1$
we have a new expression $\frac{a}{1+a^2}+\frac{b}{1+b^2}+\frac{c}{1+c^2} \longrightarrow \frac{\sqrt x}{1+x}+ \frac{\sqrt y}{1+y} +\frac{\sqrt z}{1+z } \longrightarrow f(x)+f(y)+f(z)$
Differentiating $f(z)$ twice, we get negative second derivatives for $0 \le x\le 1$, so it is concave
by Jensen's inequality, we get $ f(\frac{x+y+z}{3}) \ge \frac{f(x)}{3} +\frac{f(y)}{3}+\frac{f(z)}{3} \implies \frac{\sqrt \frac{1}{3}}{1+\frac{1}{3}}\ge \frac{f(x)}{3} +\frac{f(y)}{3}+\frac{f(z)}{3}$
$\implies \frac{3\sqrt3}{4} \ge f(x)+f(y)+f(z)= \frac{a}{1+a^2}+\frac{b}{1+b^2}+\frac{c}{1+c^2}$
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