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已知x>0,y>0,x+y=1,求证x^4+y^2>1/8
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已知x>0,y>0,x+y=1,求证x^4+y^2>1/8
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答案和解析
x=1-y
x^4+y^2
=x^4+(1-x)^2
=x^4+x^2-2x+1
=(x^2-1/2)^2+2x^2-2x+1/2
=(x^2-1/2)^2+2(x-1/2)^2+1/4
>1/4
>1/8
感觉应该是x^4+y^4≥1/8
x^4+y^4
>=[(x^2+y^2)]^2/2
>=[{(x+y)/2}^2]/2
=(x+y)^2/8
=1/8
x^4+y^2
=x^4+(1-x)^2
=x^4+x^2-2x+1
=(x^2-1/2)^2+2x^2-2x+1/2
=(x^2-1/2)^2+2(x-1/2)^2+1/4
>1/4
>1/8
感觉应该是x^4+y^4≥1/8
x^4+y^4
>=[(x^2+y^2)]^2/2
>=[{(x+y)/2}^2]/2
=(x+y)^2/8
=1/8
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