An Inequality in Triangle, with Sines II

Problem

An Inequality in Triangle, with  Sines II

Solution 1

In a triangle $\sin A,\,\sin B,\,\sin C\gt 0.\,$ In addition, $\sin A\sin B\sin C\le\displaystyle\frac{3\sqrt{3}}{8}.$

Thus applying first Hölder's inequality and, subsequently, the AM-GM inequality,

$\displaystyle \begin{align} \left(\sum_{cycl}\frac{\sin A}{\sin B}\right)\left(\sum_{cycl}\frac{\sin A}{\sin^2 B}\right)\left(\sum_{cycl}\frac{\sin A}{\sin^3 B}\right)&\ge\left(\sum_{cycl}\frac{\sin A}{\sin^2 B}\right)^3\\ &\ge\frac{27}{\sin A\sin B\sin C}\\ &\ge 27\cdot\frac{8}{3\sqrt{3}}=24\sqrt{3}. \end{align}$

Equality holds only for equilateral triangles.

Solution 2

In a triangle $\sin A,\,\sin B,\,\sin C\gt 0.\,$ In addition,

$3\sqrt{3}\ge 8\sin A\sin B\sin C\le\displaystyle\frac{3\sqrt{3}}{8}=\displaystyle\frac{abc}{R^3}.$

Thus,

$\displaystyle \begin{align} 24\sqrt{3} &\le 8R^3\left(\frac{3}{\sqrt[3]{abc}}\cdot\frac{3}{\sqrt[3]{(abc)^2}}\cdot 3\right)\\ &=8R^3\cdot 3\left(\sqrt[3]{\frac{a}{b^2}\frac{b}{c^2}\frac{c}{a^2}}\right)\cdot 3\left(\sqrt[3]{\frac{a}{b^3}\frac{b}{c^3}\frac{c}{a^3}}\right)\cdot 3\left(\sqrt[3]{\frac{a}{b}\frac{b}{c}\frac{c}{a}}\right)\\ &\le 8R^3\left(\sum_{cycl}\frac{a}{b}\right)\left(\sum_{cycl}\frac{a}{b^2}\right)\left(\sum_{cycl}\frac{a}{b^3}\right)\\ &= \left(\sum_{cycl}\frac{a/(2R)}{b/(2R)}\right)\left(\sum_{cycl}\frac{a/(2R)}{(b/(2R))^2}\right)\left(\sum_{cycl}\frac{a//(2R)}{(b/2R)^3}\right)\\ &=\left(\sum_{cycl}\frac{\sin A}{\sin B}\right)\left(\sum_{cycl}\frac{\sin A}{\sin^2 B}\right)\left(\sum_{cycl}\frac{\sin A}{\sin^3 B}\right). \end{align}$

Solution 3

With the AM-GM inequality,

$\displaystyle \begin{align} RHS &\le (3^3)\sqrt[3]{1}\left(\sqrt[3]{\frac{\displaystyle \prod_{cycl}\sin A}{\displaystyle\prod_{cycl}\sin^2B}}\right)\left(\sqrt[3]{\frac{\displaystyle \prod_{cycl}\sin A}{\displaystyle\prod_{cycl}\sin^3B}}\right)\\ &=\frac{27}{\displaystyle \prod_{cycl}\sin A}\le\frac{27\cdot 8}{3\sqrt{3}}\\ &=24\sqrt{3}. \end{align}$

Solution 4

Basically, it's all about rearrangements:

$\displaystyle \begin{align} \sum_{cycl}\frac{\sin A}{\sin B}&\ge\sum_{cycl}\frac{\sin A}{\sin A} = 3,\\ \sum_{cycl}\frac{\sin A}{\sin^2 B}&\ge\sum_{cycl}\frac{\sin A}{\sin^2 A}=\sum_{cycl}\frac{1}{\sin A},\\ &\ge\frac{3}{\sin ((A+B+C)/3)},\,\text{by Jensen's inequality},\\ &=\frac{3}{\sin (180^{\circ}/3)}= 2\sqrt{3}\\ \sum_{cycl}\frac{\sin A}{\sin^3 B}&\ge\sum_{cycl}\frac{\sin A}{\sin^3 A}=\sum_{cycl}\frac{1}{\sin^2 A}\\ &=3+\sum_{cycl}\cot^2 A\ge 4. \end{align}$

Acknowledgment

The problem from the Romanian Mathematical Magazine has been kindly posted by Dan Sitaru at the CutTheKnotMath facebook page. Solution 1 is by Kevin Soto Palacios; Solution 2 is by Myagmarsuren Yadamsuren; Solution 3 is by Soumava Chakraborty; Solution 4 is by Leo Giugiuc.

 

[an error occurred while processing this directive]

|Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny
[an error occurred while processing this directive]
[an error occurred while processing this directive]