If $a>0, b>0$, then $\frac{a}{b}+\frac{b}{a}\geq2$.

Question

Why does $a,b$ have to be positive? Clearly if we let $a=-1$ and $b=1$, then that's a counterexample. However, when I did the proof I've got to the inequality: $(a-b)^2\geq 0$, which is clearly true regardless of $a,b$ being positive. So, my question is why do both $a,b$ have to be positive?

Answer 1

Surely $a\ne0, b\ne 0$

Note that $\frac{a}{b}+\frac{b}{a}\geq2$ and $(a-b)^2\geq 0$ are NOT equivalent. Because if you want to go from $\frac{a}{b}+\frac{b}{a}\geq2$ to $(a-b)^2\geq 0$, you need to multiply $ab$ on both sides, i.e.

$$ab\cdot\left(\frac{a}{b}+\frac{b}{a}\right)\underset{?}\geq2\cdot ab$$

Note the question mark underneath the $\ge$ sign, if $ab<0$, the inequality sign will change, hence $\frac{a}{b}+\frac{b}{a}\geq2$ and $(a-b)^2\geq 0$ are NOT equivalent, and the counter-example you provided, $a=-1, b=1$ exactly shows this case.

Answer 2

Your confusion arises as a result of misinterpretation of logic. The question only states that "if (condition), then (proposition)", which has to be interpreted as if the condition is true, then the proposition follows as a consequence. It does not say anything about the inference in the other direction. To be more precise, $a > 0$ and $b > 0$ are just sufficient conditions for $(a-b)^2 \geq 0$, but not necessary conditions.

For two propositions to be equivalent, we need both sufficiency and necessity to hold.