Trigonometric and absolute value proof

Question

Prove that: $$\forall x,y\in\left(-\frac\pi2,\frac\pi2\right)\ |x+y|\le|\tan x+\tan y|$$

Answer 1

If $x,y\in\left[0,\frac{\pi}{2}\right)$, then $\tan x>x\ge0$ and $\tan y>y\ge 0$. So

$$|\tan x+\tan y|=\tan x+\tan y|>x+y=|x+y|$$

If $x,y\in\left(\frac{-\pi}{2},0\right)$, then $-x,-y\in\left(0,\frac{\pi}{2}\right)$ and hence

$$|\tan x+\tan y|=\tan (-x)x+\tan (-y)>-x-y=|x+y|$$

If $x\in\left[0,\frac{\pi}{2}\right)$ and $y\in\left(\frac{-\pi}{2},0\right)$, let $u=-y$. Then $|\tan x+\tan y|=|\tan x-\tan u|$ and $|x+y|=|x-u|$.

By symmetry, we may assume that $x\ge u$.

Then $|\tan x-\tan u|=\tan x-\tan u$ and $|x-u|=x-u$.

Let $f(x)=\tan x-u$.

$$f'(u)=\sec^2u-1=\tan^2u\ge0$$

So $f$ is increasing on $\left[0,\frac{\pi}{2}\right)$. Therefore,

$$\tan x-x\ge \tan u-u$$

$$\tan x-\tan u\ge x-u$$

$$|\tan x+\tan y|=\tan x-\tan u\ge x-u=|x+y|$$