the maximum value of $\frac{\sin A}{A}+\frac{\sin B}{B}+\frac{\sin C}{C}$

Question

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 .

Answer 1

Here's one approach to doing this:

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

Answer 2

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.