设x＞0y＞0,x≠y,x^2-y^2=x^3-y^3,求证：1＜x+y＜4/3
问题描述：
设x＞0y＞0,x≠y,x^2-y^2=x^3-y^3,求证：1＜x+y＜4/3
答
x^2-y^2=x^3-y^3
(x-y)(x+y) = (x-y)(x^2 + xy + y^2)
x≠y,所以
x + y = x^2 + xy + y^2
(x + y)^2 - (x+y) = xy
(x - y)^2 > 0
x^2 + y^2 > 2xy
x^2 + 2xy + y^2 > 4xy
(x+y)^2 > 4xy
xy (x + y)^2 - (x+y) = xy (3/4)(x+y)^2 - (x+y) (x+y)[(3/4)(x+y) - 1) (3/4)(x+y) - 1 (x+y) 另一方面
(x + y)^2 - (x+y) = xy > 0
(x + y - 1/2)^2 > 1/4
x + y - 1/2 > 1/2 和 x + y -1/2 x+y > 1 和 x+y x>0 ,y > 0 所以舍去 x + y 综上所述
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