# Show that for any positive number a and b, (a + b)/2 ≥ √ab

## Solution

### Proof.

(a + b)/2 ≥ √ab

(a + b)/2 ≥ √ab ⟺ ((a + b)/2)^{2} ≥ ab

⟺ (a + b)^{2} = 4ab

⟺ a^{2} + b^{2} + 2ab ≥ 4ab

⟺ a^{2} − 2ab + b^{2} ≥ 0

⟺ (a − b)^{2} ≥ 0

Since square on any number is greater than zero, so the (a − b)^{2} ≥ 0 and so we have the inequality.