proofs involving the triangle inequality.

Question

$$\forall a,b \in R (|a+b|=|a|+|b| \iff ab \ge 0)$$

I'm really stuck on where to even start with this. I'm assuming it has something do to with the triangle inequality, but don't know how to apply it. Here's what I can figure out anyhow.

but if $ab \ge 0$ then $a \ge 0$ and $b \ge 0$

or $a \lt 0$ and $b \lt 0$

Suppose $a \ge 0$ and $b \ge 0$

then $|a+b| \le |x| + |b|$ by the triangle inequality, but this gives an $\ge$ not just a

Suppose $a \lt 0$ and $b \lt 0$

then I have not clue what to apply in all honesty.

Can somehow point me in the right direction, please?

score 2 · Accepted Answer · answered Oct 07 '15 at 15:52

2

Hint: Square both sides of the equation $$|a + b| = |a| + |b|.$$

Can you take it from here?

answered Oct 07 '15 at 15:52

Jose Arnaldo Bebita Dris

1

$|a+b|^2=(a+b)^2=a^2+2ab+b^2$, this equals a^2+b^2 when a = 0 or b=0, but not when both are positive or negative. what else am I missing? – Osuynonma Oct 07 '15 at 16:03
Hint #2: $|x|^2 = x^2 \hspace{0.2in} \forall x$. – Jose Arnaldo Bebita Dris Oct 07 '15 at 16:05
@AlbertWolfgang, what is $(|a| + |b|)^2$? =) – Jose Arnaldo Bebita Dris Oct 07 '15 at 16:08

proofs involving the triangle inequality.

1 Answers1