Is it correct? $\tan x + \cot x \ge 2$ proof The question is to prove $\tan x + \cot x \ge 2$ when $x$ is an acute angel.
This is what I did
$$\begin{align}
\tan x + \cot x &\ge 2\\
\frac{1}{\sin x \cos x} &\ge 2\\
\left(\frac{1}{\sin x \cos x}\right) - 2 &\ge 0\\
\left(\frac{1 - 2\sin x \cos x}{\sin x \cos x}\right) &\ge 0\\
\left(\frac{(\sin x - \cos x)^2}{\sin x \cos x}\right) &\ge 0\\
\left(\frac{(\sin x - \cos x)^2}{\sin x \cos x}\right) &\ge 0\\
\end{align}$$
Both nominator and denominator will never be negative because nominator is powered to two and cosx & sinx are positive when angel is acute.
Is it correct? Is there another way to solve?

Let $ y = \tan x + \cot x $.
Then $ y' = \sec^{2} x - \csc^{2} x $.
So when $ y' = 0 $, $ x = \pi/4 $, for $x$ acute.
You can prove this is a minimum by considering $ y' $ for $ x = \pi/6 $ and $ x = \pi/3 $.
Thus $ y_{min} = \tan \pi/4 + \cot \pi/4 = 1 + 1 = 2.$
