I already proved the backward implication, and I was hoping I could get a hint for the forward. Would it be a proof by contradiction? The first part of the exercise was to prove the triangle inequality; would this be helpful here?
I am not looking for the answer, so no spoilers please. I encountered this in Kuratowski’s intro to cal, but I don’t know what tags would be appropriate, so please feel free to add them. Thanks in advance.
If you start with $|a+b| = |a| + |b|$ both sides are non-negative… hence you can square both sides… and get $ab = |ab|$
Since $|a+b|=|a|+|b|$ we get after squaring $$ |a+b|^2 =|a|^2+2|ab|+|b|^2$$Since $|x|^2=x^2$ we get $$a^2+2ab+b^2 = a^2+2|ab|+b^2$$ so $$ab =|ab|\implies ab\geq 0$$
HINT: $$|a+b|=|a|+|b|\implies a^2+2ab+b^2=|a|^2+2|a||b|+|b|^2$$ and we know that $|a|^2=a^2$...
Hint:
I'd suggest splitting the proof into cases.
If you go case analysis, you seperate them as long as it gets easier (at some point the mass of cases makes it more complicated again). So in this case, you'd test all 3 (4) cases in each assuming the sign of all variables.
Another proof can be obtained using the inner product of vectors, where $a$ and $b$ are vectors of dimension 1. Recall that $$\left|a\right|^2=a\cdot{a}$$ where '$\cdot$' is the inner product operator similar to $\left<,\right>$.
In general $$\left|a+b\right|^2=\left(a+b\right)\cdot\left(a+b\right)=\left|a\right|^2+2a\cdot{b}+\left|b\right|^2 \tag{1}$$ Now, if we assume $$\left|a+b\right|=\left|a\right|+\left|b\right|\Rightarrow\left|a+b\right|^2=\left({\left|a\right|+\left|b\right|}\right)^2=\left|a\right|^2+2\left|a\right|\left|b\right|+\left|b\right|^2 \tag{2}$$ Setting equations $(1)$ and $(2)$ equal to each other, $$\left|a+b\right|^2=\left|a\right|^2+2a\cdot{b}+\left|b\right|^2=\left|a\right|^2+2\left|a\right|\left|b\right|+\left|b\right|^2$$
Simplifying, (and dividing by 2) we are left with: $$a\cdot{b}=\left|a\right|\left|b\right|$$ This can only be true if $a\cdot{b}\geq0$, since $\left|a\right|\left|b\right|\ge0$
