Let $$P(X)=a_{0}X^{n}+a_{1}X^{n-1}+\cdots+a_{n}$$be a polynomial with real coefficients such that $a_{0}>0$ and $$a_{n}\geq a_{i}\geq 0,$$for all $i=0,1,2,\ldots,n-1.$ Prove that if $$P^{2}(X)=b_{0}X^{2n}+b_{1}X^{2n-1}+\cdots+b_{n-1}X^{n+1}+\cdots+b_{2n},$$then $P^2(1)\geq 2b_{n-1}.$
Problem
Source: Moldova TST 2017, Day 3, Problem 1
Tags: algebra
Wave-Particle
21.03.2017 00:52
I believe you have a typo -- should be $P^2(X)=b_0X^{2n}+b_1X^{2n-1}+\dots+b_{n-1}X^{n+1}+\dots+b_{2n}$, not $P^{2}(X)=b_{0}X^{2n}+b_{1}X_{2n-1}+\cdots+b_{n-1}X^{n+1}+\cdots+b_{2n}$
Snakes
21.03.2017 01:02
Wave-Particle wrote: I believe you have a typo -- should be $P^2(X)=b_0X^{2n}+b_1X^{2n-1}+\dots+b_{n-1}X^{n+1}+\dots+b_{2n}$, not $P^{2}(X)=b_{0}X^{2n}+b_{1}X_{2n-1}+\cdots+b_{n-1}X^{n+1}+\cdots+b_{2n}$ Yes sorry
countdown1000
21.03.2017 02:49
Note that $P(1)^2=(a_0+a_1+\cdots+a_n)^2$ and $b_{n-1}=a_0a_n+a_1a_{n-1}+\cdots+a_na_0$. It suffices to show that $$(a_0+a_1+\cdots+a_n)^2\ge 2a_0a_n+2a_1a_{n-1}+\cdots+2a_na_0$$However this is obvious since $(a_0+a_1+\cdots+a_n)^2=\sum_{i=0}^n a_i^2 + \sum_{i\neq j} 2a_ia_j$ and we are given that $a_i\ge 0$ for all $i$.