<s>[INST] 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?
 [/INST] Since $x$ is an acute angle, $\cot x$ and $\tan x$ are positive.
Therefore, by A.M.-G.M. inequality,
$\frac {\tan x + \cot x}{2} \ge \sqrt{\tan x \times \cot x}$
Now, $\tan x \times \cot x = 1$ (by definition)
Therefore we obtain,
$\tan x + \cot x \ge 2 \sqrt{1}$
And so, $\tan x + \cot x \ge 2 $
Proof complete.
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