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Problem 983

AMC 12 late, AIME early
Algebra Difficulty 4.5 Prove it Cesko-Slovacko-Poljsko · Czech-Polish-Slovak Mathematical Match · 2013

Show that for any real x>0x > 0 and integer n>0n > 0 we have
xn+1xn−2≥n2(x+1x−2). x^n + \frac{1}{x^n} - 2 \ge n^2 \left(x + \frac{1}{x} - 2\right).

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

Without loss of generality assume that y=x>1y = \sqrt{x} > 1. The identity
a2+1a2−2=(a−1a)2 a^2 + \frac{1}{a^2} - 2 = \left(a - \frac{1}{a}\right)^2
reduces the problem to showing that
yn−1yn≥n(y−1y) y^n - \frac{1}{y^n} \ge n\left(y - \frac{1}{y}\right)
or y2n−n(yn+1−yn−1)−1≥0y^{2n} - n(y^{n+1} - y^{n-1}) - 1 \ge 0. Upon division by y−1>0y - 1 > 0 this follows from the following computation:
y2n−1y−1−nyn−1(y2−1)y−1=∑i=02n−1yi−n(yn−1+yn)==∑i=0n−1(y2n−1−i−yn−yn−1+yi)=∑i=0n−1yi(yn−1−i−1)(yn−i−1)≥0. \begin{aligned} \frac{y^{2n} - 1}{y - 1} - \frac{n y^{n-1}(y^2 - 1)}{y - 1} &= \sum_{i=0}^{2n-1} y^i - n(y^{n-1} + y^n) = \\ &= \sum_{i=0}^{n-1} (y^{2n-1-i} - y^n - y^{n-1} + y^i) = \sum_{i=0}^{n-1} y^i (y^{n-1-i} - 1)(y^{n-i} - 1) \ge 0. \end{aligned}

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