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Problem 1727

National Olympiad, first round

Geometry Difficulty 6.4 Prove it Russian Mathematical Olympiad · Russia

Let $P(x)$ be a polynomial of degree $n \ge 2$ with nonnegative coefficients. Let $a$ , $b$ , and $c$ be the side lengths of some triangle. Prove that the numbers $\sqrt[n]{P(a)}$ , $\sqrt[n]{P(b)}$ , and $\sqrt[n]{P(c)}$ are also the side lengths of some triangle.

Пусть $P(x)$ — многочлен степени $n \ge 2$ с неотрицательными коэффициентами. Пусть $a$ , $b$ , $c$ — длины сторон некоторого треугольника. Докажите, что числа $\sqrt[n]{P(a)}$ , $\sqrt[n]{P(b)}$ , $\sqrt[n]{P(c)}$ также являются длинами сторон некоторого треугольника.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Assume that $a \ge b \ge c$ . Set $G(x) = P(x)/x^n$ and notice that $G(a) \le G(b) \le G(c)$ .

Let, without loss of generality, $a \ge b \ge c$ ; these three positive numbers are the side lengths of a triangle if and only if $a < b + c$ . Since the coefficients of $P(x)$ are nonnegative, we have $P(a) \ge P(b) \ge P(c) > 0$ ; thus, we need to check that $\sqrt[n]{P(a)} < \sqrt[n]{P(b)} + \sqrt[n]{P(c)}$ .

Let $P(x) = p_n x^n + p_{n-1} x^{n-1} + \dots + p_0$ . Denote $G(x) = P(x)/x^n$ . Note that
$G(a) = p_n + \frac{p_{n-1}}{a} + \dots + \frac{p_0}{a^n} \le p_n + \frac{p_{n-1}}{b} + \dots + \frac{p_0}{b^n} = G(b)$
and similarly, $G(a) \le G(c)$ . Therefore,
$\sqrt[n]{P(a)} = a \sqrt[n]{G(a)} < (b + c) \sqrt[n]{G(a)} \le b \sqrt[n]{G(b)} + c \sqrt[n]{G(c)} = \sqrt[n]{P(b)} + \sqrt[n]{P(c)},$
which is what was required to prove.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.