Let $P_n(x)=a_0 + a_1x + \cdots + a_nx^n$, with $n \geq 2$, be a real-coefficient polynomial. Prove that if there exists $a > 0$ such that \begin{align*} P_n(x) = (x + a)^2 \left( \sum_{i=0}^{n-2} b_i x^i \right), \end{align*}where $b_i$ are positive real numbers, then there exists some $i$, with $1 \leq i \leq n-1$, such that \[a_i^2 - 4a_{i-1}a_{i+1} \leq 0.\]
Problem
Source: China TST 2002 Quiz
Tags: algebra, polynomial, inequalities, algebra unsolved
ywl
05.07.2009 18:46
Can anyone help? I've been stuck on this problem. It doesn't seem easy.
Batominovski
22.06.2013 00:43
Well, I don't understand the problem statement.
chaotic_iak
22.06.2013 09:18
I think the last words of the problem is wrong. It should be "...$i = 0, 1, \ldots, n-2$, then there is some $i$..."
xxp2000
23.06.2013 21:04
chaotic_iak wrote: I think the last words of the problem is wrong. It should be "...$i = 0, 1, \ldots, n-2$, then there is some $i$..." We assume $a=1$ WLOG. For $a\ne1$, we only need to look at $\frac{P(ax)}{a^n}$. Obviously we have $a_i=b_{i-2}+2b_{i-1}+b_i,i=0,\cdots,n$, where we let $b_i=0$ for $i<0$ or $i>n-2$. Suppose $a_i^2-4a_{i-1}a_{i+1}>0,\forall 0<i<n$. $(b_{i-1}+\frac{b_{i-2}+b_i}2)^2>(b_{i-1}+2b_{i-2})(b_{i-1}+2b_i)$ implies $\frac{(b_i+b_{i-2})^2}4>(b_i+b_{i-2})b_{i-1}$, or $b_{i-1}<\frac{b_i+b_{i-2}}4$ for $0<i<n$. So $b_1>4b_0$, $b_i>4b_{i-1}-b_{i-2},i=2,\cdots,n-2$ and $b_{n-2}<\frac{b_{n-3}}4$. The first two inequalities imply $b_i-(2+\sqrt{3})b_{i-1}>0,i=1,\cdots,n-2$. Contradiction with the third inequality. Q.E.D.