<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] Even if your implications are correct, your logic is flawed.
Logically, what you have done is similar to
$$\begin{align}
+1&=-1
\\ \Rightarrow (+1)^2&=(-1)^2
\\ \Rightarrow 1&=1.
\end{align}$$
Now $1=1$ so $+1=-1$ must be true also.
Deriving a true statement says nothing about the truth of the hypothesis.
</s>