MathDB

1

Part of 1995 Taiwan National Olympiad

Problems(1)

$P(x)=a_0+a_1x+...+a_nx^n\in\mathbb{C}[x]$

Source: 4-th Taiwanese Mathematical Olympiad 1995

1/16/2007
Let P(x)=a0+a1x+...+anxn∈C[x]P(x)=a_{0}+a_{1}x+...+a_{n}x^{n}\in\mathbb{C}[x] , where an=1a_{n}=1. The roots of P(x)P(x) are b1,b2,...,bnb_{1},b_{2},...,b_{n}, where ∣b1∣,∣b2∣,...,∣bj∣>1|b_{1}|,|b_{2}|,...,|b_{j}|>1 and ∣bj+1∣,...,∣bn∣≤1|b_{j+1}|,...,|b_{n}|\leq 1. Prove that ∏i=1j∣bi∣≤∣a0∣2+∣a1∣2+...+∣an∣2\prod_{i=1}^{j}|b_{i}|\leq\sqrt{|a_{0}|^{2}+|a_{1}|^{2}+...+|a_{n}|^{2}}.
inequalitiesalgebrapolynomialalgebra proposed