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Question

Real numbers x, y, and z satisfy the following equality: 4(x + y + z) = x^2 + y^2 + z^2.Let M be the maximum value of xy + xz + yz, and let m be the minimum value of xy + xz + yz. Find M + 10m.

Accepted Answer

Let A = x + y + z, B = x^2 + y^2 + z^2, and C = xy + xz + yz. We are told that 4A = B.Then A^2 = (x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + xz + yz) = B + 2C = 4A + 2C.Hence, C = 12 (A - 2)^2 - 2.Also, B - C = x^2 + y^2 + z^2 - (xy + xz + yz) = (x - y)^2 + (x - z)^2 + (y - z)^22 0,so C B. Then A^2 = B + 2C 3B = 12A. Hence, 0 A 12, so -2 C 48. We see that C = -2 when (x,y,z) = (2,-2,2), and C = 48 when (x,y,z) = (4,4,4), so M = 48 and m = -2, and M + 10m = 28.