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

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

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

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

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

  • 3
    $\begingroup$ Well, the $\implies$ arrows all go in the wrong direction. Also it looks like if this were correct it would show that actually $|a+b|=|a|+|b|$... $\endgroup$ – David C. Ullrich Jul 18 '15 at 19:00
  • 2
    $\begingroup$ Your last step assumes $\sqrt{a^2}\sqrt{b^2}=ab$. It doesn't. It is $|ab|$. Still since ab\leq |ab|$ the last step works. You still have the problem that your implications go in the wrong direction. $\endgroup$ – Thomas Andrews Jul 18 '15 at 19:02
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    $\begingroup$ You can't start assuming what you want to prove. Also, concluding something true in the end does not prove anything. Let me prove that $1=2$. Well, multiply both sides by $0$, so $0=0$. Hence I conclude that $1=2$. $\endgroup$ – Ivo Terek Jul 18 '15 at 19:03
  • $\begingroup$ If I point my arrows in the opposite direction, would that then constitute a correct proof of this inequality ? $\endgroup$ – Mutating Algorithm Jul 18 '15 at 19:08

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,

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


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



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