Dan Sitaru's Inequality with Tangents


Dan Sitaru's inequality with tangents in a triangle

Proof 1

The starting point is Heinz's inequality:

For $x,y\gt 0,\;$ $\alpha\in [0,1],$

$x^{1-\alpha}y^{\alpha}+x^{\alpha}y^{1-\alpha}\le x+y.$

We apply Heinz's inequality with $\displaystyle\alpha=\frac{1}{3}:

$x^{\frac{2}{3}}y^{\frac{1}{3}}+x^{\frac{1}{3}}y^{\frac{2}{3}}\le x+y$

and set $x=\tan A\;$ and $y=\tan B\;$ to obtain

$\displaystyle\sqrt[3]{\tan A}\sqrt[3]{\tan B}(\sqrt[3]{\tan A}+\sqrt[3]{\tan B})\le \tan A+\tan B.$

Similarly we get

$\displaystyle\sqrt[3]{\tan B}\sqrt[3]{\tan C}(\sqrt[3]{\tan B}+\sqrt[3]{\tan C})\le \tan B+\tan C,\\ \displaystyle\sqrt[3]{\tan C}\sqrt[3]{\tan A}(\sqrt[3]{\tan C}+\sqrt[3]{\tan A})\le \tan C+\tan Q.$

Adding all three and recollecting that in a triangle

$\tan A+\tan B+\tan C=\tan A\tan B\tan C.$

yields the required inequality.

Proof 2

Let $\tan^{\frac{1}{3}}(A)=a;\;$ $\tan^{\frac{1}{3}}(B)=b,\;$ and $\tan^{\frac{1}{3}}(C)=c.\;$ In any triangle $A +B+ C=\pi;\;$ therefore, $\tan A+ \tan B+ \tan C=\tan A\tan B\tan C,\;$ i.e., $a^3+ b^3+ c^3=a^3b^3c^3.\;$ After the substitution we have to prove that $\displaystyle \sum_{cycl}ab(a+ b)\le 2a^3b^3c^3.\;$ Now,

$\displaystyle\begin{align} \sum_{cycl}ab(a+ b)&\le\sum \frac{a^2 +b^2}{2}(a+ b)\\ &=\sum_{cycl}a^3+ \sum_{cycl}\frac{ab(a +b)}{2}\\ \end{align}$

It follows that

$\displaystyle\begin{align} \sum_{cycl} ab( a+ b)&\le 2\sum_{cycl}a^3\\ &=2a^3b^3c^3. \end{align}$


Dan Sitaru has kindly posted the problem, with a solution (Proof 1), at the CutTheKnotMath facebook page. Proof 2 is by Ritesh Dutta (India).


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

Copyright © 1996-2017 Alexander Bogomolny


Search by google: