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 .

By canceling the gradient of
$$\frac{\sin A}A+\frac{\sin B}B+\frac{\sin(A+B)}{\pi-A-B},$$
we must have
$$\frac{A\cos A-\sin A}{A^2}=-\frac{(\pi-A-B)\cos(A+B)+\sin(A+B)}{(\pi-A-B)^2}=\frac{B\cos B-\sin B}{B^2}.$$
As the function
$$\frac{x\cos x-\sin x}{x^2}$$ is monotonic in the first quadrant, we have $A=B$, and we now maximize
$$2\frac{\sin A}A+\frac{\sin2A}{\pi-2A}.$$
Graphically, it is obvious that the function has a single maximum, and we can verify that $A=\dfrac\pi3$ cancels the derivative. It would be better to prove that there are no other solutions.
