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P^2(1) \geq 2b_{n-1} [Moldova TST 2017, D3, P1]

Source: Moldova TST 2017, Day 3, Problem 1

March 20, 2017
algebra

Problem Statement

Let P(X)=a0Xn+a1Xn−1+⋯+anP(X)=a_{0}X^{n}+a_{1}X^{n-1}+\cdots+a_{n} be a polynomial with real coefficients such that a0>0a_{0}>0 and an≥ai≥0,a_{n}\geq a_{i}\geq 0, for all i=0,1,2,…,n−1.i=0,1,2,\ldots,n-1. Prove that if P2(X)=b0X2n+b1X2n−1+⋯+bn−1Xn+1+⋯+b2n,P^{2}(X)=b_{0}X^{2n}+b_{1}X^{2n-1}+\cdots+b_{n-1}X^{n+1}+\cdots+b_{2n}, then P2(1)≥2bn−1.P^2(1)\geq 2b_{n-1}.
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