I was wondering if there is perhaps a way to check whether an inequality is true or false without trying to brute force a counter-example or trying to actually prove the statement first? Perhaps an indication that can be proved useful?
Also, upon trying to prove the first inequality, I couldn't seem to prove the first statement? Is it because it's false, or am I missing something?
Here's the question
That's what I have tried doing (for part a)
$\frac{(1-a)(1-b)}{ab} \ge 1$
$(1-a)(1-b) \ge ab$
$1-b-a \ge 0$
However, this doesn't really help me much.
Thank you!
Best Answer
Let $a,b>0$. Note that the following are equivalent:
$$\frac{1-a}{a}\cdot\frac{1-b}{b}\ge1$$ $$(1-a)(1-b)\ge ab$$ $$1-a-b+ab\ge ab$$ $$1-a-b\ge0$$ $$a+b\le1$$
Hence $\frac{1-a}{a}\cdot\frac{1-b}{b}\ge1$ holds iff $a+b\le1$ holds. We can use this to show that (a) is true and (b) is false.
So for (a) let $a+b\le\frac{1}{2}$. It follows that $a+b\le1$. Hence $\frac{1-a}{a}\cdot\frac{1-b}{b}\ge1$.
For (b), we need to pick $a$, $b$, with $\frac{1}{2}<a+b\le1$. So let $a=b=\frac{1}{3}$. Then $\frac{1-a}{a}\cdot\frac{1-b}{b}=4\ge1$, but $a+b=\frac{2}{3}>\frac{1}{2}$.