Given the positive numbers $a, b, c$. Prove that $\frac{a}{\sqrt{a^2+1}}+\frac{b}{\sqrt{b^2+1}}+\frac{c}{\sqrt{c^2+1}}\le \frac{3}{2}$ Given the positive numbers $a, b, c$ satisfy $a+b+c\le \sqrt{3}$. Prove that $\frac{a}{\sqrt{a^2+1}}+\frac{b}{\sqrt{b^2+1}}+\frac{c}{\sqrt{c^2+1}}\le \frac{3}{2}$ 
My Try (Edited from Comments): 
By Cauchy Schwarz, we have that $$ (a^2+1)(1+3) \ge \left(a+\sqrt {3}\right)^2 \rightarrow \frac{a}{\sqrt{a^2+1}} \le \frac{2a}{a+\sqrt{3}} $$  I need a new method

Note that $f(x)=\frac{x}{\sqrt{x^2+1}}$ is a concave function for $x>0$. This follows as $$f''(x)=-\dfrac{3x}{(x^2+1)^{\frac{5}{2}}}<0$$ Now use Jensen's inequality. Note that $$f(a)+f(b)+f(c) \le 3 f\left(\frac{a+b+c}{3}\right)=3f\left(\frac{\sqrt{3}}{3}\right)=\frac{3}{2}$$
As $a+b+c=\sqrt{3}$. 
The maximum is achieved when $a=b=c$. We are done!
