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Consider the comparison between (a + b)p and a p + bp, where a, b and p are positive. It is elementary that (a + b)p > a p + bp for p > 1 and the opposite holds for 0 < p < 1 (let b/a = x ≤ 1: then for p > 1, we have (1 + x)p > 1 + x > 1 + x p). Let us write Fp(a, b) = (a + b)p − a p − bp. For p = 2, 3, we have the identities F2(a, b) = 2ab, F3(a, b) = 3(a2b + ab2). Also, when b/a is small, (a + b)p is approximated by a p + pa p−1b. These facts suggest that it is a natural idea to look for estimates of Fp(a, b) in terms of G p(a, b) = a p−1b + abp−1.

Inequalities comparing $(a+b)^p−a^p−b^p$ and $a^{p−1}b+ab^{p−1}$