An Inequality with Arctangents in Triangle

Problem

An Inequality with Arctangents in Triangle

Proof 1

By direct verification, $\displaystyle\arctan\frac{1}{2}+\arctan\frac{1}{5}+\arctan\frac{1}{8}=\frac{\pi}{4}.\,$ Thus, by Chebyshev's inequality,

$\displaystyle\begin{align} \text{LHS} &= \frac{36}{\pi}\left(\sum_{cycl}a^3\right)\left(\sum_{cycl}\cos^3A\right)\\ &\ge \frac{4}{3\pi}\left(\left(\sum_{cycl}a\right)\left(\sum_{cycl}\frac{\cos A}{3}\right)\right)^3\\ &= \frac{32s^3}{3\pi}\left(\sum_{cycl}\frac{\cos A}{3}\right)^3\\ &= \text{RHS}. \end{align}$

Proof 2

By Hölder's inequality,

$\begin{align}&\left(\displaystyle\frac{a^3\cos^3A}{\arctan\frac{1}{2}}+\frac{b^3\cos^3B}{\arctan\frac{1}{5}}+\frac{c^3\cos^3C}{\arctan\frac{1}{8}}\right)\\ &\qquad\times\left(\arctan\frac{1}{2}+\arctan\frac{1}{5}+\arctan\frac{1}{8}\right)\\ &\qquad\times(1+1+1)\\ &\qquad\ge (a\cos A+b\cos B+c\cos C)^3. \end{align}$

By direct verification, $\displaystyle\arctan\frac{1}{2}+\arctan\frac{1}{5}+\arctan\frac{1}{8}=\frac{\pi}{4}.\,$ It follows that

$\displaystyle\begin{align} \text{LHS} &\ge \frac{4R(\sin 2A+\sin 2B+\sin 2c)}{3\pi}\\ &=\frac{4(4R\sin A\sin B\sin C)^3}{3\pi}\\ &= \frac{\displaystyle 4\left(4R\frac{rs}{2R^2}\right)^3}{3\pi}\\ &= \text{RHS}. \end{align}$

Proof 3

By Radon's Inequality

$\displaystyle\frac{a^3\cos^3A}{\arctan\frac{1}{2}}+\frac{b^3\cos^3B}{\arctan\frac{1}{5}}+\frac{c^3\cos^3C}{\arctan\frac{1}{8}}\\ \qquad\displaystyle\ge\frac{(a\cos A+b\cos B+c\cos C)^3}{\left(\sqrt{\arctan\frac{1}{2}}+\sqrt{\arctan\frac{1}{5}}+\sqrt{\arctan\frac{1}{8}}\right)^2}.$

Further,

(1)

$\begin{align} &a\cos A+b\cos B+c\cos C\\ &\qquad\qquad = R(\sin 2A+\sin 2B+\sin 2C)\\ &\qquad\qquad = 4R\sin 2A\sin 2B\sin 2C\\ &\qquad\qquad = 4R\frac{abc}{8R^3}\\ &\qquad\qquad = \frac{2\Delta}{R}, \end{align}$

where $\Delta\,$ is the area of triangle $ABC.\,$ Next

(2)

$\begin{align} &\left(\sqrt{\arctan\frac{1}{2}}+\sqrt{\arctan\frac{1}{5}}+\sqrt{\arctan\frac{1}{8}}\right)^2\\ &\qquad\qquad \le 3\left(\arctan\frac{1}{2}+\arctan\frac{1}{5}+\arctan\frac{1}{8}\right)\\ &\qquad\qquad = 3\frac{\pi}{4}. \end{align}$

Combining (1) and (2),

$\displaystyle\begin{align} &\frac{(a\cos A+b\cos B+c\cos C)^3}{\left(\sqrt{\arctan\frac{1}{2}}+\sqrt{\arctan\frac{1}{5}}+\sqrt{\arctan\frac{1}{8}}\right)^2}\\ &\qquad\qquad\ge\frac{8\Delta^2}{R^3}\cdot\frac{4}{3\pi}\\ &\qquad\qquad\frac{32\cdot r^2\cdot s^2}{3\pi R^3}. \end{align}$

Acknowledgment

Dan Sitaru has kindly posted the above problem from his book Math Accent, with two proofs, at the CutTheKnotMath facebook page. Proof 1 is by Anas Adlany; Proof 2 is by Kevin Soto Palacios; Proof 3 is by Myagmarsuren Yadamsuren.

 

|Contact| |Front page| |Contents| |Algebra| |Store|

Copyright © 1996-2017 Alexander Bogomolny

 62050231

Search by google: