Prove that $|a+b| \le |a|+|b|$

Question

I got this one from Demidovich's "Problems and exercises in Calculus".

I tried to rewrite this as

$$\sqrt{(\sqrt{a^2}-\sqrt{b^2})^2} \le \sqrt{(a-b)^2} \\ \Leftrightarrow \sqrt{a^2-2\sqrt{(ab)^2}+b^2} \le \sqrt{a^2-2ab+b^2} \\ \Leftrightarrow a^2-2ab+b^2 \le a^2-2ab+b^2$$

This certainly proves that they are equal but one isn't lesser than the other... is it correct? It looks like something is missing. If it's correct, then why $\le$ and not "="?

Answer 1

This is the famous triangle inequality. The standard proof for this is

$$-|x|\leq x\leq |x|$$

$$-|y|\leq y\leq |y|$$

Then adding these gives

$$-(|x|+|y|)\leq x+y\leq |x|+|y|$$

Now, use the fact that for any real numbers $a$ and $b$

$$|b|\leq a\Leftrightarrow -a\leq b\leq a$$

to get

$$|x+y|\leq |x|+|y|$$