Maths Olympiad Prep

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Algebra Difficulty 6.1 National olympiad Prove it

(2) If nn is odd, prove: AH⩽−n−2n+2(n−1)n(AG)n.(−1997\frac{A}{H} \leqslant-\frac{n-2}{n}+\frac{2(n-1)}{n}\left(\frac{A}{G}\right)^{n} .(-1997 Korean Mathematical Olympiad problem)

Solution

(2) If n⩾3n \geqslant 3 is odd
AH⩽(AG)n=(AG)n+n−2n[(AG)n−1]=−n−2n+2(n−1)n(AG)n\frac{A}{H} \leqslant\left(\frac{A}{G}\right)^{n}=\left(\frac{A}{G}\right)^{n}+\frac{n-2}{n}\left[\left(\frac{A}{G}\right)^{n}-1\right]=-\frac{n-2}{n}+\frac{2(n-1)}{n}\left(\frac{A}{G}\right)^{n}

It can be proven that the proposition is independent of the parity of nn.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.