Mongolian Mathematical Olympiad

Question

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) 1 P'(0). (1) Prove that 0 Q(x) x P(x) for all 0 x 1. (2) Prove that P(Q(x)) Q(P(x)) for all 0 x 1. It is *not* necessary to study the conditions for equality.

Accepted Answer

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 x 1. (1) For x = 0, we have Q(0) = 0 = P(0). For 0 < x 1, we have 0 Q(x)/x Q(1)/1 1 P'(0) P(x)/x, thus 0 Q(x) x P(x). (2) If Q(x) = 0, we have P(Q(x)) = P(0) = 0 Q(P(x)). For Q(x) > 0, we have P(Q(x))/Q(x) P(x)/x and Q(x)/x Q(P(x))/P(x), therefore P(Q(x)) P(x)Q(x)/x Q(P(x)).