Find the minimum value of $a \tan^2 x + b \cot^2 x,$ where $\text{ a is greater than b, b is greater than 0}.$

Question

Find the minimum value of $a \ tan^2 x + b \cot^2 x.\text{ a is greater than b, b is greater than 0}$. Closest thing to solution I can come up with is this $(\sqrt{a} \ tan x + \sqrt{b}\cot x )^2 -2 \sqrt{a}\sqrt{b} $

Answer 1

Let $t=(\tan \, x)^{2}$. Note that $at+\frac b t =(\sqrt {at} -\sqrt {\frac b t})^{2}+2\sqrt {ab} \geq 2\sqrt {ab}$. Also equality holds when $t=\sqrt {b/a}$. Hence the minimum value is $2\sqrt {ab}$ which is attained when $\tan \, x=(b/a)^{1/4}$.

Answer 2

The AM-GM inequality says: $a\tan^2x + b\cot^2x \ge 2\sqrt{a\tan^2x\cdot b\cot^2x}= 2\sqrt{ab}$. What does this mean...?

Answer 3

You are almost there:

$(√a|\tan x| -√b|\cot x|)^2 +$

$2√a√b|\tan x||\cot x| \ge$

$ 2√a√b|\tan x \cot x| =2√a√b.$

Equality:

$√a|\tan x| -√b|\cot x| =0.$

$\tan ^2 x =√b/√a.$