# Is this a valid proof for $\left | a+b \right | \leq \left |a \right| + \left | b \right |$?

$\left | a+b \right | \leq \left |a \right| + \left | b \right | \Rightarrow$

$\sqrt{{(a+b)}^2} \leq \sqrt{{a}^2} + \sqrt{{b}^2}$

${(\sqrt{{(a+b)}^2})}^2 \leq ({\sqrt{{a}^2} + \sqrt{{b}^2}})^2 \Rightarrow$

${(a+b)}^2 \leq {a}^2 + 2\sqrt{{a}^2}\sqrt{{b}^2} + {b}^2 \Rightarrow$

${a}^2 + 2ab + {b}^2 \leq {a}^2 + 2ab + {b}^2$ , This is true since $\left | x \right| \leq \left |x \right | \forall x \epsilon \mathbb{R}$

$\therefore \left | a+b \right | \leq \left |a \right| + \left | b \right |$

I'm not sure if the the last statement in my manipulation is enough to prove the original inequality.

• Well, the $\implies$ arrows all go in the wrong direction. Also it looks like if this were correct it would show that actually $|a+b|=|a|+|b|$... – David C. Ullrich Jul 18 '15 at 19:00

QED!
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