设x＞0y＞0,x≠y,x^2-y^2=x^3-y^3,求证：1＜x+y＜4/3

题目

设x＞0y＞0,x≠y,x^2-y^2=x^3-y^3,求证：1＜x+y＜4/3

答案

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 /4
(x + y)^2 - (x+y) = xy < (x+y)^2 /4
(3/4)(x+y)^2 - (x+y) < 0
(x+y)[(3/4)(x+y) - 1) < 0
(3/4)(x+y) - 1 < 0
(x+y) < 4/3
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 < -1/2
x+y > 1 和 x+y < -1
x>0 ,y > 0 所以舍去 x + y < -1
综上所述
1 < x+y < 4/3