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

$$\begin{align}
\tan x + \cot x &amp;\ge 2\\
\frac{1}{\sin x \cos x} &amp;\ge 2\\
\left(\frac{1}{\sin x \cos x}\right) - 2 &amp;\ge 0\\
\left(\frac{1 - 2\sin x \cos x}{\sin x \cos x}\right) &amp;\ge 0\\
\left(\frac{(\sin x - \cos x)^2}{\sin x \cos x}\right) &amp;\ge 0\\
\left(\frac{(\sin x - \cos x)^2}{\sin x \cos x}\right) &amp;\ge 0\\
\end{align}$$

Both nominator and denominator will never be negative because nominator is powered to two and cosx &amp; sinx are positive when angel is acute.

Is it correct? Is there another way to solve?
using the lemma $$t+\frac{1}{t}\geq 2$$ for $t&gt;0$  we get the inequality $$\tan(x)+\frac{1}{\tan(x)}\geq 2$$
your way is also ok.
Even if your implications are correct, your logic is flawed.

Logically, what you have done is similar to

$$\begin{align}
+1&amp;=-1
\\ \Rightarrow (+1)^2&amp;=(-1)^2
\\ \Rightarrow 1&amp;=1.
\end{align}$$
Now $1=1$ so $+1=-1$ must be true also.

Deriving a true statement says nothing about the truth of the hypothesis.