Olympiad Maths Prep

Library / /1 of 6

Algebra Difficulty 4.8 AIME Prove it Belarus

Given a polynomial P(x)P(x) with positive real coefficients.
Prove that P(1)P(xy)P(x)P(y)P(1)P(xy) \ge P(x)P(y) for all x1,y1x \ge 1, y \ge 1.

Solution

We use weighted Chebyshev's inequality: if a1,,an>0a_1, \dots, a_n > 0, and x1x2xnx_1 \le x_2 \le \dots \le x_n, y1y2yny_1 \le y_2 \le \dots \le y_n, then
(a1x1++anxn)(a1y1++anyn)a1++ana1x1y1++anxnyn.() \frac{(a_1x_1 + \dots + a_nx_n)(a_1y_1 + \dots + a_ny_n)}{a_1 + \dots + a_n} \le a_1x_1y_1 + \dots + a_nx_ny_n. \quad (*)

Now let p(x)=a1+a2x++anxn1p(x) = a_1 + a_2x + \dots + a_nx^{n-1} be the given polynomial. Then we simply set in ()(*) x1=1x_1 = 1, x2=xx_2 = x, ..., xi=xi1x_i = x^{i-1}, y1=1y_1 = 1, ..., yi=yi1y_i = y^{i-1}, thus finishing the proof.

Looking for a route rather than an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.