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Sunday, April 17, 2016

For real numbers \displaystyle 0\lt x\lt \frac{\pi}{2}, prove that \cos^2 x \cot x+\sin^2 x \tan x\ge 1. (Second Solution)

For real numbers \displaystyle 0\lt x\lt \frac{\pi}{2}, prove that \cos^2 x \cot x+\sin^2 x \tan x\ge 1.

My solution:

First, note that we can rewrite \cos^2 x \cot x+\sin^2 x \tan x as \displaystyle \frac{\cos^3 x}{\sin x}+\frac{\sin^3 x}{\cos x}.

For the domain \displaystyle 0\lt x\lt \frac{\pi}{4}, we have \cos^3 x\gt \sin^3x,\,\dfrac{1}{\sin x}\gt \dfrac{1}{\cos x}, so by the rearrangement inequality we have:

\begin{align*}\cos^2 x \cot x+\sin^2 x \tan x=\dfrac{\cos^3 x}{\sin x}+\dfrac{\sin^3 x}{\cos x}&\ge \dfrac{\cos^3 x}{\cos x}+\dfrac{\sin^3 x}{\sin x}\\&=\cos^2x+\sin^2 x\\&=1\end{align*}

By the same token, for the domain \displaystyle \frac{\pi}{4}\le x\lt \frac{\pi}{2}, we have \sin^3 x\gt \cos^3x,\,\dfrac{1}{\cos x}\gt \dfrac{1}{\sin x}, so by the rearrangement inequality we also have:

\begin{align*}\cos^2 x \cot x+\sin^2 x \tan x=\dfrac{\sin^3 x}{\cos x}+\dfrac{\cos^3 x}{\sin x}&\ge \dfrac{\sin^3 x}{\sin x}+\dfrac{\cos^3 x}{\cos x}\\&=\sin^2x+\cos^2 x\\&=1\end{align*}

Combining the two yields the result.

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