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Algebra Difficulty 4.5 AIME Prove it Soviet Union

Problem:

Show that for any real numbers x,y>1x, y > 1, we have
x2y−1+y2x−1≥8. \frac{x^2}{y - 1} + \frac{y^2}{x - 1} \geq 8.

Solution

Solution:

We have (x−2)2≥0(x - 2)^2 \geq 0, so x2≥4(x−1)x^2 \geq 4(x - 1). Hence,
xx−1≥2. \frac{x}{\sqrt{x - 1}} \geq 2.
Now by AM/GM,
x2y−1+y2x−1≥2xy(x−1)(y−1). \frac{x^2}{y - 1} + \frac{y^2}{x - 1} \geq \frac{2xy}{\sqrt{(x - 1)(y - 1)}}.
But the right-hand side ≥2⋅2⋅2=8\geq 2 \cdot 2 \cdot 2 = 8.

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