Nov. 28, 2020
Archangel Macsika
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
⟺ a2 + b2 + 2ab ≥ 4ab
⟺ a2 − 2ab + b2 ≥ 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.
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