# How to prove $\dfrac{\sin(A)}{A} +\dfrac{\sin(B)}{B}+\dfrac{\sin(C)}{C}< \dfrac{9*3^{0.5}}{2\pi}$

Only for an acute angle triangle. $A$,$B$,$C$ are angles of a triangle. This isnt sine rule form. Ive tried Cauchy Schwarz theorem , A.M, G.M form but am unable to get the above result. Could someone point me in the right direction ? In an equilater triangle the sign becomes =

• Is the right side of that inequality supposed to be $\dfrac{9\sqrt{3}}{2\pi}$? Commented Oct 30, 2014 at 17:52
• Yes that is correct Commented Oct 30, 2014 at 17:54
• I believe the question is that $\alpha$, $\beta$ and $\gamma$ are angles in radians and $\alpha$ is thus, not the length of the side opposite $\alpha$... (for those who are thinking like that). Commented Oct 30, 2014 at 18:01
• $\pi/2-1$ is not acute.
– user175968
Commented Oct 30, 2014 at 18:31

Outline: First, show that $f(x) = \dfrac{\sin x}{x}$ is concave down over $x \in [0,\frac{\pi}{2})$.
Then, apply Jensen's inequality to get $\dfrac{f(A)+f(B)+f(C)}{3} \le f\left(\dfrac{A+B+C}{3}\right)$.
• Any hint on how to show it's concave down? $f''(x)=-\frac{(x^2-2)\sin(x)+2xcos(x)}{x^3}$, so we need to prove $(x^2-2)sin(x)+2xcos(x)\geq0$ for x on $[0,\pi/2)$ which is not that trivial.
• @frank000 It's equivalent to $\cot x > \frac{2-x^2}{2x}$. When the RHS is negative, we're good; otherwise it's equivalent to $\tan x < \frac{2x}{2-x^2}$. Set $x=\sqrt2\tan\theta$; now it's equivalent to $\tan(\sqrt2\tan\theta) < \frac1{\sqrt2}\tan(2\theta)$. This follows from the convexity of tan (in the form $\tan(\lambda u)\le\lambda\tan u$ when $\lambda\in[0,1]$, used twice).