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# Understand the problem

Find all the polynomials $P(x)$ of a degree $\leq n$ with real non-negative coefficients such that $P(x) \cdot P(\frac{1}{x}) \leq [P(1)]^2$ , $\forall x>0$.
##### Source of the problem
Albanian BMO TST 2009
Algebra
Easy
##### Suggested Book
An Excursion in Mathematics

Do you really need a hint? Try it first!

This problem is all about non-negative real numbers. The first thing that should come to your mind is “standard inequalities!”.
Write $P(x)=\Sigma a_kx^k$. Using the Cauchy-Schwarz inequality, show that $P(x)P(1/x)\ge (P(1))^2$.
Note that hint 2 along with the hypothesis in the problem implies that $P(x)P(1/x)=(P(1))^2$. Hence equality holds in hint 2.
As equality holds in CS, it means that for all $k$ satisfying $a_k\neq 0$, $\frac{a_kx^k}{a_kx^{-k}=x^{2k}$ is a constant. This is absurd, hence there can be at most one such $k$. Hence, only monomials can satisfy the given inequality.

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