discrete maths direct proof

Question

16. Prove that if a and b are positive integers, then a/b + b/a ≥ 2.

Answer 1

It's important to recall that for any arbitrary integer n, we must have n²≥0. With that said, if a and b are positive integers, then a-b is an integer and so it follows that (a-b)²≥0. Foiling out the left hand side yields

(a-b)²≥0 ⇒ a²-2ab+b²≥0 ⇒ a²+b²≥2ab.

Finally, we can divide both sides by ab (since ab ≠ 0 because a and b are positive integers) to get the following:

a²+b²≥2ab ⇒ a²/(ab)+b²/(ab)≥2 ⇒ a/b+b/a≥2.

This completes the proof.

Answer 2

a/b + b/a = (a² + b²) / ab

(a-b)² >= 0

So a² - 2ab + b² >= 0

So a² + b² >= 2ab

as a and b are positive, so is ab

^{Divide both sides by ab and you get the required result}