Consider three real numbers $a$, $b$, and $c$. Prove that we can pick two of them such that their product is non-negative.
Using proof by cases:
$a >0,\space b>0,\space c>0$
Above, we can see that if we pick ANY two pairs that their product is alway positive.
Ex: $$a\cdot b = ab \qquad (ab>0)$$
Lets say $a <0, b<0,c>0$
There is only one way to get a positive product out of this group. You need to pick two variables that have the same "sign"(The variables must both be positive or negative). If the two variables are not the same sign then their product will always be negative.
Ex: $$a \cdot b = ab \qquad (ab>0, a<0, b<0)$$ $$a \cdot c = ac \qquad (ac <0, a<0, b >0)$$
$\therefore$ You can pick two variables with like signs from this group and their producgt will always be positive. $\square$
Is this the correct way to go about this proof? I don't see another way with the small set of proof methods we have. I don't know if this is enough to prove this though. It seems too simple... Any thoughts?