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

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 $f’(1/3)=3\sqrt{3}/16$, 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.
