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


As $A+B+C=\pi$

$C=\pi -(A+B)$

After differentiating it


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 .

  • $\begingroup$ What is dA dB dC $\endgroup$
    – qwr
    Commented Jul 25, 2018 at 7:13
  • $\begingroup$ @qwr After differentiating it . $\endgroup$
    – Koolman
    Commented Jul 25, 2018 at 7:48
  • 1
    $\begingroup$ wrt what variable? $\endgroup$
    – qwr
    Commented Jul 25, 2018 at 7:50
  • $\begingroup$ Well, it is a perfectly legal and rigorous expression, as long as it is interpreted as a differential. $\endgroup$
    – lisyarus
    Commented Jul 25, 2018 at 8:26
  • $\begingroup$ Similar: math.stackexchange.com/q/874630/42969. $\endgroup$
    – Martin R
    Commented Jul 25, 2018 at 8:38

2 Answers 2


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.)


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.


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