If a and B are positive numbers, and a ≠ B, the proof is A2B + B2A > A + B

It is proved that: A2B + B2A − (a + b) = A3 + B3 − A2B − ab2ab = A2 (a − b) − B2 (a − b) AB = (A2 − B2) (a − b) AB = (a + b) (a − b) 2Ab. ∵ a and B are positive numbers, ∵ a + b > 0, AB > 0, and (a-b) 2 > 0, so (a + b) (a − b) 2Ab > 0, that is, A2B + B2A − (a + b) > 0, ∵ A2B

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