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Problem 905

AMC 12 late, AIME early

Algebra Difficulty 4.7 Prove it Berkeley Math Circle: Monthly Contest 5 · United States

Let $P$ be a polynomial with positive real coefficients. Prove that if
$P\left(\frac{1}{x}\right) \geq \frac{1}{P(x)}$
holds for $x=1$ , then it holds for every $x>0$ .

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.

Official solution

Solution:
Let $P(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0}$ , so $a_{k}>0$ for every $k$ . Since the statement holds for $x=1$ , $P(1) \geq 1$ . Then by Cauchy-Schwarz,
$P(x) P\left(\frac{1}{x}\right)=\left(\sum_{k=0}^{n}\left(\sqrt{a_{k} x^{k}}\right)^{2}\right)\left(\sum_{k=0}^{n}\left(\sqrt{\frac{a_{k}}{x^{k}}}\right)^{2}\right) \geq\left(\sum_{k=0}^{n} a_{k}\right)^{2}=P(1)^{2} \geq 1$

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