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

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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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