An Inequality with Tangents and Cotangents

Problem

An Inequality with Tangents and Cotangents

Solution 1

Set $\displaystyle\tan\frac{A}{2}\tan\frac{B}{2}=z,\,$ $\displaystyle\tan\frac{B}{2}\tan\frac{C}{2}=x,\,$ $\displaystyle\tan\frac{C}{2}\tan\frac{A}{2}=y.\,$ We have to prove that

$\displaystyle\left(x+\frac{1}{x}\right)\left(y+\frac{1}{y}\right)\left(z+\frac{1}{z}\right)\ge\frac{1000}{27},$

subject to $x+y+z=1.$

Consider function $f:\;(0,1)\rightarrow\mathbb{R},\,$ defined by $\displaystyle f(t)=\ln\left(t+\frac{1}{t}\right).\,$ We find $\displaystyle f'(t)=\frac{2t}{t^2+1}-\frac{1}{t}\lt 0,\,$ $\displaystyle f''(t)=\frac{2(1-t^2)}{(t^2+1)^2}+\frac{1}{t^2}\gt 0,\,$ for $t\in (0,1),\,$ implying that the function is convex and decreasing so that, via Jensen's inequality,

$\displaystyle\begin{align} f(x)+f(y)+f(z) &\ge 3f\left(\frac{x+y+z}{3}\right)\\ &=3f\left(\frac{1}{3}\right), \end{align}$

i.e.

$\displaystyle\ln\left(x+\frac{1}{x}\right)+\ln\left(y+\frac{1}{y}\right)+\ln\left(z+\frac{1}{z}\right)\ge 3\ln\left(3+\frac{1}{3}\right)=\ln\left(\frac{10}{3}\right),$

implying the required inequality.

Solution 2

With the same change of variables, the problem reduces to

$\displaystyle\left(x+\frac{1}{x}\right)\left(y+\frac{1}{y}\right)\left(z+\frac{1}{z}\right)\ge\frac{1000}{27},$

subject to $x+y+z=1.$

With Hölder's inequality, we get

$\displaystyle\left(x+\frac{1}{x}\right)\left(y+\frac{1}{y}\right)\left(z+\frac{1}{z}\right)\ge\left(\sqrt[3]{xyz}+\frac{1}{\displaystyle\sqrt[3]{xyz}}\right)^3.$

By the AM-GM inequality, $\displaystyle\sqrt[3]{xyz}\le\frac{1}{3},\,$ and, since the function $\displaystyle f(t)=t+\frac{1}{t}\,$ is decreasing for $t\in (0,1),\,$

$\displaystyle \left(\sqrt[3]{xyz}+\frac{1}{\displaystyle\sqrt[3]{xyz}}\right)^3\ge \left(3+\frac{1}{3}\right)^3=\frac{1000}{27},$

which completes the proof.

Acknowledgment

I am grateful to Dan Sitaru for communicating to me his problem and its two solutions by Leo Giugiuc.

 

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