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
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}$
$\implies \frac{3\sqrt3}{4} \ge f(x)+f(y)+f(z)= \frac{a}{1+a^2}+\frac{b}{1+b^2}+\frac{c}{1+c^2}$
Substitute $a^2, b^2, c^2$ by $x, y, z$. Then we will try to bound $\sqrt{x}/(1+x)$ by a linear form of $x$. Consider a line tangent to the expression when the equality occurs i.e. when $x=1/3$. Evaluating $fâ(1/3)=3\sqrt{3}/16$, we find that such line is $y=\frac{3\sqrt{3}(x+1)}{16}$. By simple algebraic manipulation, $$\frac{\sqrt{x}}{1+x}\leq\frac{3\sqrt{3}(x+1)}{16}$$ is equivalent to $(3x-1)^2(3x^2+14x+27)\geq0$, which is obviously true. The desired result follows immediately.
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 $a^2+b^2+c^2=1.$
