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Mongolian Mathematical Olympiad

Algebra Difficulty 5.0 AIME, harder Prove it Mongolia

Let aa, bb, cc be distinct positive real numbers. Show that
(b−c)4(a−b)2(a−c)2+(a−c)4(a−b)2(b−c)2+(a−b)4(a−c)2(b−c)2≥332. \frac{(b-c)^4}{(a-b)^2(a-c)^2} + \frac{(a-c)^4}{(a-b)^2(b-c)^2} + \frac{(a-b)^4}{(a-c)^2(b-c)^2} \ge \frac{33}{2}.

Solution

Let us assume that a>b>ca > b > c. a−c=(a−b)+(b−c)≥2(a−b)(b−c)a - c = (a - b) + (b - c) \ge 2\sqrt{(a - b)(b - c)}. Then
(a−c)4(a−b)2(b−c)2≥16 \frac{(a - c)^4}{(a - b)^2 (b - c)^2} \ge 16
holds. Now it suffices to prove that
(a−b)4(b−c)2+(b−c)4(a−b)2≥(c−a)22 \frac{(a-b)^4}{(b-c)^2} + \frac{(b-c)^4}{(a-b)^2} \geq \frac{(c-a)^2}{2}
By Cauchy-Schwarz inequality
(a−b)4(b−c)2+(b−c)4(a−b)2≥((a−b)2+(b−c)2)2(b−c)2+(a−b)2=(a−b)2+(b−c)2 \frac{(a-b)^4}{(b-c)^2} + \frac{(b-c)^4}{(a-b)^2} \geq \frac{((a-b)^2 + (b-c)^2)^2}{(b-c)^2 + (a-b)^2} = (a-b)^2 + (b-c)^2
Then again by Cauchy-Schwarz (a−b)2+(b−c)2≥(a−b+b−c)22=(a−c)22(a-b)^2 + (b-c)^2 \geq \frac{(a-b+b-c)^2}{2} = \frac{(a-c)^2}{2}. Equality holds for a+c=2ba+c=2b.

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