Prove that for all rational number $a$ and $b$, $\frac{a+b}{2}$ ≥ $\sqrt{ab}$

Question

Prove that for all rational numbers (can be integer) a y b, $\frac{a+b}{2}$ ≥ $\sqrt{ab}$

https://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means — T.J. Gaffney, Feb 15 '18 at 18:11
The inequality holds for appropriate real numbers, not merely rational numbers. — David G. Stork, Feb 15 '18 at 18:13

score 1 · Answer 1 · answered Feb 15 '18 at 18:11

1

Hint:

$$(a+b)^2\ge4ab\iff (a-b)^2\ge0$$

It is important $\;a,b\;$ non-negative ...

answered Feb 15 '18 at 18:11

DonAntonio

score 0 · Answer 2 · answered Feb 15 '18 at 18:12

0

by squaring and rearranging we get $$(a-b)^2\geq 0$$ which is true.

answered Feb 15 '18 at 18:12

2 Answers2