Maths Olympiad Prep

Track / Stage 5 / 21 of 400 #621 of 1964

Problem 621

AIME late
Algebra Difficulty 5.0 Prove it Iranian Mathematical Olympiad · Iran

Suppose that xx, yy and zz are positive real numbers and x2+y2+z2=x2y2+y2z2+z2x2x^2 + y^2 + z^2 = x^2 y^2 + y^2 z^2 + z^2 x^2. Prove that
(x−y)2(y−z)2(z−x)2≤(x2−y2)2+(y2−z2)2+(z2−x2)2. (x - y)^2 (y - z)^2 (z - x)^2 \le (x^2 - y^2)^2 + (y^2 - z^2)^2 + (z^2 - x^2)^2.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

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Official solution

Because of the problem's assumption, it is enough to prove that
(∏(x−y))2(∑x2)≤∑(x2−y2)2(∑(xy)2). (\prod (x-y))^2 (\sum x^2) \le \sum (x^2 - y^2)^2 (\sum (xy)^2).
By Cauchy-Schwarz inequality we have
(∑xy(x2−y2))2≤∑(x2−y2)2(∑(xy)2).(1) (\sum xy(x^2 - y^2))^2 \le \sum (x^2 - y^2)^2 (\sum (xy)^2). \quad (1)
On the other hand, an easy calculation shows that
(∏(x−y))2(∑x)2=(∑xy(x2−y2))2. (\prod (x-y))^2 (\sum x)^2 = (\sum xy(x^2 - y^2))^2.
Finally, we have
(∏(x−y))2(∑x2)≤(∏(x−y))2(∑x)2=(∑xy(x2−y2))2.(2) (\prod (x-y))^2 (\sum x^2) \le (\prod (x-y))^2 (\sum x)^2 = (\sum xy(x^2 - y^2))^2. \quad (2)

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.