Polynomials
(All-Russian Mathematical Olympiad 2007 Grade 10 Day 1 P2, AoPS, proposed by A. Khrabrov) Consider a polynomial \(P(x) = a_0 x^n + a_1 x^{n - 1} + \dots + a_{n - 1}x + a_n\) with integer coefficients. Let \(m\) denote the minimum of the integers \[ a_0, a_0 + a_1, \dots, a_0 + a_1 + \dots + a_n . \] Prove that \(P(x) \geq m x^n\) for all \(x \geq 1\).
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Note that
\(\begin{align*} P(x) & = a_0 x^n + a_1 x^{n - 1} + \dots + a_{n - 1}x + a_n \\ & = a_0 (x^n - x^{n - 1}) + (a_0 + a_1)(x^{n - 1} - x^{n - 2}) \\ & \quad + \dots + (a_0 + a_1 + \dots + a_{n - 1})(x - 1) + (a_0 + a_1 + \dots + a_n) \end{align*} \\)
holds, which implies that
\(\begin{align*} & P(x) - m x^n \\ & = a_0 (x^n - x^{n - 1}) + (a_0 + a_1)(x^{n - 1} - x^{n - 2}) \\ & \quad + \dots + (a_0 + a_1 + \dots + a_{n - 1})(x - 1) + (a_0 + a_1 + \dots + a_n) \\ & \quad - m (x^n - x^{n - 1}) - m (x^{n - 1} - x^{n - 2}) - \dots - m (x - 1) - m \\ & = (a_0 - m)(x^n - x^{n - 1}) + (a_0 + a_1 - m)(x^{n - 1} - x^{n - 2}) \\ & \quad + \dots + (a_0 + a_1 + \dots + a_{n - 1} - m)(x - 1) + (a_0 + a_1 + \dots + a_n - m) . \end{align*} \\)
It follows that for any \(x \geq 1\), \[ P(x) \geq m x^n . \]
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