Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 5.6 AIME, harder Prove it Mongolia

Let $P(x)$ and $Q(x)$ be polynomials with non-negative real coefficients, and let $P'(x)$ denote the derivative of $P(x)$ . Suppose that we have $P(0) = Q(0) = 0$ and $Q(1) \le 1 \le P'(0)$ .
(1) Prove that $0 \le Q(x) \le x \le P(x)$ for all $0 \le x \le 1$ .
(2) Prove that $P(Q(x)) \le Q(P(x))$ for all $0 \le x \le 1$ .
It is not necessary to study the conditions for equality.

Solution

Since $P(0) = Q(0) = 0$ and the coefficients of $P(x)$ and $Q(x)$ are non-negative, we see that the functions $P(x)/x$ and $Q(x)/x$ are increasing for $x > 0$ .
Let $0 \le x \le 1$ .
(1) For $x = 0$ , we have $Q(0) = 0 = P(0)$ . For $0 < x \le 1$ , we have
$0 \le Q(x)/x \le Q(1)/1 \le 1 \le P'(0) \le P(x)/x,$
thus $0 \le Q(x) \le x \le P(x)$ .

(2) If $Q(x) = 0$ , we have $P(Q(x)) = P(0) = 0 \le Q(P(x))$ . For $Q(x) > 0$ , we have $P(Q(x))/Q(x) \le P(x)/x$ and $Q(x)/x \le Q(P(x))/P(x)$ , therefore
$P(Q(x)) \le P(x)Q(x)/x \le Q(P(x)).$

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