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?

It is quite easy...
$\tan x + \cot x$
= $\tan x + \cot x - 2 \sqrt {\tan x \cot x} + 2 \sqrt{\tan x \cot x}$ 
=$(\sqrt {\tan x} -\sqrt {\cot x})^2+ 2$   [because $\tan x \cot x =1$]

Since square of any number is greater than or equal to zero, we can write, 
$(\sqrt {\tan x} -\sqrt {\cot x})^2 \ge 0$ 
So, $(\sqrt {\tan x} -\sqrt {\cot x})^2 + 2 \ge 2$
But, $(\sqrt {\tan x} -\sqrt {\cot x})^2 + 2 = \tan x + \cot x$ 
So, $\tan x + \cot x \ge 2$
NOTE : We want any square number to make this problem easier, this is a standard technique for proving most of inequalities in mathematics, therefore we have added and subtract $2 \sqrt{\tan x \cot x}$  in the main expression to make the expression perfect square.
