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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

enter image description here

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!

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    $\begingroup$ How about 2a less than or equal to 1-b and similarly for 2b and substituting it back. $\endgroup$
    – danny
    Commented Jan 6, 2020 at 2:03
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    $\begingroup$ If $a+b\le\dfrac12$ then $1-a-b\ge\dfrac12$, so certainly $1-b-a\ge0$ (which is what you got when you tried) $\endgroup$
    – J. W. Tanner
    Commented Jan 6, 2020 at 2:06
  • $\begingroup$ I totally missed it! So, I can basically write it as such: $a + b ≤ \frac{1}{2} ≤ 1$, which is obviously true? $\endgroup$
    – Flavio Esposito
    Commented Jan 6, 2020 at 2:09
  • $\begingroup$ Also, is there a generally accepted way of checking whether an inequality is true or false without having to brute force a counter-example to disprove it? $\endgroup$
    – Flavio Esposito
    Commented Jan 6, 2020 at 2:13
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    $\begingroup$ Note that what you did in your attempt was an equivalent rewriting. $\endgroup$
    – A.Γ.
    Commented Jan 6, 2020 at 2:19

1 Answer 1

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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 $a+b=\frac{2}{3}>\frac{1}{2}$.

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  • $\begingroup$ Thank you! Just to clarify, to find a counter-example for part b you had to use the given information (If $a+b ≤ \frac{1}{2}$) in the question, and the information found in part a ($a+b ≤ 1$)? Was that the reason you could disprove the second inequality in part b? $\endgroup$
    – Flavio Esposito
    Commented Jan 6, 2020 at 2:24
  • $\begingroup$ I'm just confused about how you knew that part b is false. Did you just randomly pick a number between $1$ and $\frac{1}{2}$ that worked in disproving the inequality in part b? Thank you! $\endgroup$
    – Flavio Esposito
    Commented Jan 6, 2020 at 2:27
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    $\begingroup$ We showed that $\frac{1-a}{a}\cdot\frac{1-b}{b}\ge1$ would hold if and only if $a+b\le1$. So if we can find $a$, $b$ with $\frac{1}{2}<a+b\le1$, then $\frac{1-a}{a}\cdot\frac{1-b}{b}\ge1$ will hold, but $a+b\le\frac{1}{2}$ will not hold. So we just had to find $a$, $b$ with $\frac{1}{2}<a+b\le1$. So I picked $\frac{1}{3}$ for both $a$ and $b$. $\endgroup$
    – user729424
    Commented Jan 6, 2020 at 2:27
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    $\begingroup$ Thanks, that makes perfect sense! However, what if it was the case we started with trying to prove the second inequality first? How would we know it's false? Would we be expected to try and come up with a counter-example and hope it's right, or try to prove it and see that it's impossible? $\endgroup$
    – Flavio Esposito
    Commented Jan 6, 2020 at 2:35
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    $\begingroup$ I think it's helped to start with $\frac{1-a}{a}\cdot\frac{1-b}{b}$, since this inequality is "more complicated" than $a+b\le1$, so there's more that you can do with it. Since $a,b>0$, we can start with $\frac{1-a}{a}\cdot\frac{1-b}{b}\ge1$, and then get $(1-a)(1-b)\ge ab$. From here we get $1-a-b\ge0$ and then $a+b\le1$. Also, it's really helpful to realize that each of these steps is reversible. Hence if we start with $\frac{1-a}{a}\cdot\frac{1-b}{b}\ge1$ then we get $a+b\le1$, and if we start with $a+b\le1$, we get $\frac{1-a}{a}\cdot\frac{1-b}{b}\ge1$. $\endgroup$
    – user729424
    Commented Jan 6, 2020 at 2:54

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