若已知x的立方加y的立方等于二,证明x加y小于二

设：x+y>2
则：x^3+y^3=(x+y)(x^2+y^2-xy)
>2[(x+y)^2-3xy]
>2(4-3xy)
≥2(4-3*(x+y)^2/4)
>2(4-3*4/4)
=2(4-3)
=2
即：x^3+y^3>2
与x^3+y^3=2矛盾
所以
x+y=