Two different answers by applying AM GM Inequality

Question

Find minimum value of $f(x)=\sec^2 x+\csc ^2x$

We have $$f(x)=2+\tan^2 x+\cot ^2 x$$

Method $1.$ we have $$\frac{\tan^2 x+\cot^2 x}{2} \ge 1$$ $\implies$

$$f(x) \ge 4$$

Method $2.$

We have $$\frac{2+\tan^2 x+\cot^2 x}{3} \ge 2^{\frac{1}{3}}$$ $\implies$

$$f(x) \ge 3(2^{\frac{1}{3}})-2$$

But whats wrong in method $2$?

Answer 1

AM=GM holds only in case all terms are equals but $\tan x=\cot x=2$ is impossible.

Answer 2

The second equality is untenable as $\tan x=\cot x\implies\tan x=?$

Answer 3

Because in your second AM-GM the equality should be occur for $2=\tan^2x=\cot^2x,$ which is impossible.

The right way is the following.

By AM-GM $$f(x)\geq2+2\sqrt{\tan^2x\cot^2x}=2+2=4.$$ The equality occurs for $x=45^{\circ},$ which says that $4$ is a minimal value.