已知xy为任意实数,求证x^4+y^4≥1/2xy(x+y)^2
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已知xy为任意实数,求证x^4+y^4≥1/2xy(x+y)^2
2(x^4+y^4)-xy(x+y)^2
=2x^4+2y^4-x^3y-2x^2y^2-xy^3
=(x^4-2x^2y^2+y^4)+(x^4+y^4-x^3y-xy^3)
=(x^2-y^2)^2+x^3(x-y)+y^3(y-x)
=(x^2-y^2)^2+(x-y)(x^3-y^3)
=(x^2-y^2)^2+(x-y)[(x-y)(x^2+xy+y^2)]
=(x^2-y^2)^2+(x-y)^2(x^2+xy+y^2)
=(x^2-y^2)^2+(x-y)^2[(x^2+xy+y^2/4)+3y^2/4]
=(x^2-y^2)^2+(x-y)^2[(x+y/2)^2+3y^2/4]
因为(x^2-y^2)^2≥0
(x-y)^2≥0
(x+y/2)^2+3y^2/4≥0
所以(x^2-y^2)^2+(x-y)^2[(x+y/2)^2+3y^2/4]≥0
所以2(x^4+y^4)-xy(x+y)^2≥0
所以x^4+y^4≥1/2xy(x+y)^2
=2x^4+2y^4-x^3y-2x^2y^2-xy^3
=(x^4-2x^2y^2+y^4)+(x^4+y^4-x^3y-xy^3)
=(x^2-y^2)^2+x^3(x-y)+y^3(y-x)
=(x^2-y^2)^2+(x-y)(x^3-y^3)
=(x^2-y^2)^2+(x-y)[(x-y)(x^2+xy+y^2)]
=(x^2-y^2)^2+(x-y)^2(x^2+xy+y^2)
=(x^2-y^2)^2+(x-y)^2[(x^2+xy+y^2/4)+3y^2/4]
=(x^2-y^2)^2+(x-y)^2[(x+y/2)^2+3y^2/4]
因为(x^2-y^2)^2≥0
(x-y)^2≥0
(x+y/2)^2+3y^2/4≥0
所以(x^2-y^2)^2+(x-y)^2[(x+y/2)^2+3y^2/4]≥0
所以2(x^4+y^4)-xy(x+y)^2≥0
所以x^4+y^4≥1/2xy(x+y)^2
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