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, 2012

Algebra Difficulty 4.5 AIME Prove it Belarus

Given $0 < a < b < c$ prove that
$a^{20}b^{12} + b^{20}c^{12} + c^{20}a^{12} < b^{20}a^{12} + a^{20}c^{12} + c^{20}b^{12}.$

Solution

Lemma. Let $0 < x < y < z$ and $\lambda > 1$ . Then $\sum_{cyc} x^{\lambda}y < \sum_{cyc} x y^{\lambda}$ .

Proof. Consider the function $f(y) = \sum_{cyc} (x^{\lambda}y - x y^{\lambda})$ . It takes the value $0$ for $y = x$ and $y = z$ . We have also $f''(y) = \lambda(\lambda - 1)y^{\lambda}(z-x) > 0$ , hence $f$ is convex. It follows that $f(y) < 0$ for $y \in (x; z)$ . The lemma is proved.

Now use the lemma for $x = a^{12}$ , $y = b^{12}$ , $z = c^{12}$ , and $\lambda = 20/12$ thus obtaining the inequality needed.

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