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?

We already have that
$$\tan x + \cot x = \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} = \frac{\sin^2 x + \cos^2 x}{\sin x \cos x} = \frac{1}{\sin x \cos x}$$
If $x$ is an acute angle strictly between $0$ and $\pi/2$, then $\sin x \cos x > 0$. Hence $\tan x + \cot x \geq 2$ if and only if $$1 \geq 2\sin x \cos x \ \ \ \ \  \text{  or alternatively }  \ \ \ \ \ \  \sin(2x) \leq 1$$
and this last result is true because $\sin\theta \leq 1$ for all angles $\theta$.

In other words, you almost had it. In your second line, you just had to realize that $$2\sin x \cos x$$ could be written as $\sin 2x$ and then bounded above by $1$.
