the maximum value of $\frac{\sin A}{A}+\frac{\sin B}{B}+\frac{\sin C}{C}$ 
For any acute angled triangle ABC , find the maximum value of $\frac{\sin A}{A}+\frac{\sin B}{B}+\frac{\sin C}{C}$ .

Attempt: 
As $A+B+C=\pi$
$C=\pi -(A+B)$
After differentiating it
$dA+dB+dC=0$
Now :
$\frac{\sin A}{A}+\frac{\sin B}{B}+\frac{\sin C}{C}$
$\frac{\sin A}{A}+\frac{\sin B}{B}+\frac{\sin (A+B)}{\pi-(A+B)}$
$(\frac{A\cos A-\sin A}{A^2})dA + (\frac{B\cos B- \sin B}{B^2})dB + (\frac{C\cos C-\sin c}{C^2})dC =0$
But could not solve further .

Here's one approach to doing this:


*

*Take $f(x)=\frac{\sin x}{x}$ and show that $f''(x)<0$ for $x\in(0,\pi/2)$.

*Use Jensen's inequality to conclude that for any $A,B,C\in(0,\pi/2)$ with $A+B+C$ fixed, $f(A)+f(B)+f(C)$ is maximised when they are all equal.


(Further hint for 1: write $f''(x)$ as a fraction and differentiate the numerator.)
