Dear Uncle Colin,

I have to prove that, if $a > b \ge 2$, then $ab > a + b$. I can see it must be true, but I can’t prove it! Any ideas?

Questioning Everything Done

Hi, QED, and thanks for your message!

There are several ways to go about this, as usual. I’m going to talk through two: a nugget-hunt, and a direct proof. Let’s do it directly, to put our minds at rest.

### A direct proof

The idea in the direct proof is to say that $a$ is a bit more than 2, and $b$ is 2 and possibly a bit more. If we work with those extra bits, knowing neither is negative and at least one is positive, the result drops out.

Let $a = 2 + x$ and $b = 2 + y$, where $x > 0$ and $y \ge 0$.

Then $ab = 4 + 2(x+y) + xy$ and $a+b = 4 + x + y$.

Since $2(x+y) > (x+y)$ and $xy \ge 0$, $ab > a+b \blacksquare$

### A nugget-hunt

That’s a bit pat, though. In reality, producing a proof usually has two phases: first you convince yourself it’s true by fiddling around and ignoring some of the rules, then you write it out formally to convince everyone else.

For example, you might start from what you’re trying to prove, and juggle the inequality around: what if you make it $ab - a > b$? You could factorise the left-hand side to get $a(b-1) > b$, or $a > \frac{b}{b-1}$. Is that obviously true? It’s true, but not “obvious” obvious - the right-hand inequality is 2 or smaller. Can we prove that $\frac{b}{b-1} \le 2$ if $b \ge 2$?

Look at $b \le 2(b-1)$, which is the same as $2 \le b$ - so we’re on to something! Let’s work carefully backwards:

• $2 \le b$
• $0 \le b-2$
• $b \le 2b - 2$
• $b \le 2(b-1)$
• So $\frac{b}{b-1} \le 2$ (as long as $b-1 > 0$, which it is).
• We also know $a > 2$, so $a \gt \frac{b}{b-1}$
• So $a(b-1) \gt b$ (as long as $b-1 > 0$, which it is).
• Then $ab - a \gt b$ and $ab \gt a + b \blacksquare$

There are other nuggets you could find. For example, you might rewrite the target as $ab - a - b > 0$, then think “I could factorise that if there was an extra 1” and say the target is $(a-1)(b-1) > 1$, which is clearly true because both factors are at least 1 and at least one of them is strictly greater. You can then start from “Consider…”

• Consider $(a-1)(b-1)$ with $a > 2$ and $b \ge 2$
• Clearly, $(a-1)(b-1) > 1$
• So $ab - a - b + 1 > 1$
• And $ab > a+b \blacksquare$.

Hope that helps!

- Uncle Colin

* Edited 2019-12-06 because I got my $a$s and $b$s and $x$s and $y$s and a bit of LaTeX mixed up. Thanks, Adam, for pointing it out!