Given the positive numbers $a, b, c$. Prove that $\frac{a}{\sqrt{a^2+1}}+\frac{b}{\sqrt{b^2+1}}+\frac{c}{\sqrt{c^2+1}}\le \frac{3}{2}$

Question

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

Answer 1

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!

Answer 2

By AM-GM and P-M we obtain: $$\sum_{cyc}\frac{a}{\sqrt{a^2+1}}=\sum_{cyc}\frac{a}{\sqrt{a^2+3\cdot\frac{1}{3}}}\leq\sum_{cyc}\frac{a}{\sqrt{4\sqrt[4]{\frac{a^2}{27}}}}= \frac{3\sqrt[8]{27}}{2}\frac{\sum\limits_{cyc}\sqrt[4]{a^3}}{3}\leq$$ $$\leq\frac{3\sqrt[8]{27}}{2}\left(\frac{a+b+c}{3}\right)^{\frac{3}{4}}\leq\frac{3\sqrt[8]{27}}{2}\left(\frac{\sqrt3}{3}\right)^{\frac{3}{4}}=\frac{3}{2}$$

Answer 3

Another way.

Since $ab+ac+bc\leq\frac{1}{3}(a+b+c)^2\leq1$, by AM-GM we obtain: $$\sum_{cyc}\frac{a}{\sqrt{1+a^2}}\leq\sum_{cyc}\frac{a}{\sqrt{ab+ac+bc+a^2}}=\sum_{cyc}\frac{a}{\sqrt{(a+b)(a+c)}}\leq$$ $$\leq\frac{1}{2}\sum_{cyc}\left(\frac{a}{a+b}+\frac{a}{a+c}\right)=\frac{1}{2}\sum_{cyc}\left(\frac{a}{a+b}+\frac{b}{a+b}\right)=\frac{3}{2}.$$ Done!