jmc

Question

Let x and y be two distinct positive real numbers. We define three sequences (A_n), (G_n), and (H_n) as follows. First, A_1, G_1, and H_1 are the arithmetic mean, geometric mean, and harmonic mean of x and y, respectively. Then for n 2, A_n, G_n, H_n are the arithmetic mean, geometric mean, and harmonic mean of A_n - 1 and H_n - 1, respectively. Consider the following statements: 1. A_1 > A_2 > A_3 > . 2. A_1 = A_2 = A_3 = . 4. A_1 < A_2 < A_3 < . 8. G_1 > G_2 > G_3 > . 16. G_1 = G_2 = G_3 = . 32. G_1 < G_2 < G_3 < . 64. H_1 > H_2 > H_3 > . 128. H_1 = H_2 = H_3 = . 256. H_1 < H_2 < H_3 < . Enter the labels of the statements that must hold. For example, if you think the statements labeled 2, 8, and 64 are true, enter 2 + 8 + 64 = 74.

Accepted Answer

By AM-GM-HM, A_1 G_ 1 H_1.Since x and y are distinct, equality cannot occur, so A_1 > G_1 > H_1. Note that G_1 = xy, and A_1 H_1 = x + y2 21x + 1y = x + y2 4xyx + y = xy,so G_1^2 = A_1 H_1. Now, suppose A_n > G_n > H_n for some positive integer n, and that G_n^2 = A_n H_n. Then by AM-GM-HM, A_n + 1 > G_n + 1 > H_n + 1. Also, A_n + 1 = A_n + H_n2 < A_n + A_n2 = A_n.Also, G_n + 1 = A_n H_n = G_n,and H_n + 1 = 21A_n + 1H_n > 21H_n + 1H_n = H_n.Also, by the same calculation as above, we can verify that G_n + 1^2 = A_n + 1 H_n + 1. Then by induction, we can say that A_n + 1 < A_n, G_n + 1 = G_n, H_n + 1 > H_nfor all positive integers n. Hence, the statements that are true are 1, 16, and 256, and their sum is 273.