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

Question

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

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

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

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

${a}^2 + 2ab + {b}^2 \leq {a}^2 + 2ab + {b}^2$ , 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.

Answer 1

Look at this:

Assuming $a, b \in \Bbb R$,

$a^2 + 2ab + b^2 = a^2 + 2ab + b^2, \tag{1}$

certainly true; also,

$a^2 = \vert a \vert^2; b^2 = \vert b \vert^2;$ $(a + b)^2 = \vert a + b \vert^2; ab \le \vert a \vert \vert b \vert; \tag{2}$

all also certainly true; thus,

$\vert a + b \vert^2 = (a + b)^2 = a^2 + 2ab + b^2$ $\le \vert a \vert^2 + 2\vert a \vert \vert b \vert + \vert b \vert^2 = (\vert a \vert + \vert b \vert)^2, \tag{3}$

whence

$\vert a + b \vert \le \vert a \vert + \vert b \vert. \tag{4}$

QED!