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?

Nice Work
Just don't start by assuming the equality or inequality


Indeed there is another way, There's an elementary way

Since we are only concerned about acute angles
$$\left(a-\frac1a\right)^2\ge0$$
$$a^2-2+\frac1{a^2}\ge0$$
$$a^2+\frac1{a^2}\ge2$$
put $a^2=\tan x$
$$\tan x+\frac1{\tan x}\ge2$$
$$\tan x+\cot x\ge2$$
$\text{Game Over!}$
