ISL 2017 A1

Problem A1: Let a_1, a_2, \dots, a_n, k and M be positive integers such that

    \[ \frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_n} = k \qquad \textrm{and} \qquad a_1 a_2 \cdots a_n = M. \]

If M > 1, prove that the polynomial

    \[ P(x) = M(x + 1)^k - (x + a_1)(x + a_2) \cdots (x + a_n) \]

has no positive roots.

Solution A1: We appeal to a similar strategy that is used in IMO 2012-2. Write

    \[ x + a_i = (x + 1) + 1 + \cdots + 1 \geq a_i (x + 1)^{\frac{1}{a_i}}. \]

In particular, for x \geq -1, this implies

    \[ (x + a_1) \cdots (x + a_n) \geq M (x + 1)^k, \]

so x \geq -1 \implies P(x) \leq 0 with equality only at x = 0. Thus x > 0 \implies P(x) < 0, which is equivalent to P(x) has no positive roots.

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