## Sunday, April 17, 2016

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

For real numbers [MATH]0\lt x\lt \frac{\pi}{2}[/MATH], 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 [MATH]\frac{\cos^3 x}{\sin x}+\frac{\sin^3 x}{\cos x}[/MATH].

For the domain [MATH]0\lt x\lt \frac{\pi}{4}[/MATH], 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 [MATH]\frac{\pi}{4}\le x\lt \frac{\pi}{2}[/MATH], 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.