# A geometric inequality concerning angles

Prove or disprove that: In an acute angled triangle if $A,B,C$ are angles $a,b,c,$ are respective opposite sides and $R$ is circum-radius, then
$$\prod_{cyc}\bigg(\dfrac{2A}{\pi}\bigg)^{\dfrac 1a} \le \bigg(\dfrac{2}{3}\bigg)^{\dfrac{\sqrt 3}{R}}$$

I tried to use AM-GM after representing the $a$ as $\sin$, but didn't succeed. Please help.

• You should say that you have an equality in the case of an equilateral triangle. – Jean Marie May 5 '17 at 22:04

I think it's true for all triangle.

Indeed, we need to prove that $$\sum_{cyc}\frac{1}{a}\ln\frac{2\alpha}{\pi}\leq\frac{\sqrt3}{R}\ln\frac{2}{3}$$ or $$\sum_{syc}\frac{\ln\frac{2\alpha}{\pi}}{\sin\alpha}\leq2\sqrt3\ln\frac{2}{3}.$$ Let $f(x)=\frac{\ln\frac{2x}{\pi}}{\sin{x}}.$ We see that $f''(x)<0$ for all $x\in(0,1.752...)$,

which says that $f$ is a concave function for all $x\in\left(0,\frac{\pi}{3}\right]$.

Thus, by the Vasc's LCF Theorem it's enough to prove our inequality for $\beta=\alpha$.

Hence, $\gamma=\pi-2\alpha$, where $0<\alpha<\frac{\pi}{2}$ and we need to prove that $$\frac{2\ln\frac{2\alpha}{\pi}}{\sin\alpha}+\frac{\ln\frac{2(\pi-2\alpha)}{\pi}}{\sin2\alpha}\leq2\sqrt3\ln\frac{2}{3},$$ which is indeed true.

Done!

About LCF Theorem see here: https://diendantoanhoc.net/index.php?app=core&module=attach&section=attach&attach_id=4416

• Sir can you please suggest some books for inequalities? I also like solving inequalities but have very little exposure to them. – user428700 May 6 '17 at 5:35
• @DarthVader1056 I think the best book it's the last book of Vasile Cirtoaje (he wrote it two years ago). – Michael Rozenberg May 6 '17 at 5:51