如果x,y,z是不相等的正数,证明(x^2y^2+y^2z^2+z^2x^2)/(x+y+z)>=xyz
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如果x,y,z是不相等的正数,证明(x^2y^2+y^2z^2+z^2x^2)/(x+y+z)>=xyz
P=x^2y^2+y^2z^2=y^2(x^2+z^2)>=2(y^2)xz
Q=y^2z^2+z^2x^2=z^2(y^2+x^2)>=2(z^2)xy
H=x^2y^2+z^2x^2=x^2(y^2+z^2)>=2(x^2)yz
要证(x^2y^2+y^2z^2+z^2x^2)/(x+y+z)>=xyz
即证(x^2y^2+y^2z^2+z^2x^2)>=(x^2)yz+(y^2)xz+(z^2)xy
又知P+Q+H=2(x^2y^2+y^2z^2+z^2x^2)>=2[(x^2)yz+(y^2)xz+(z^2)xy]
得证.
Q=y^2z^2+z^2x^2=z^2(y^2+x^2)>=2(z^2)xy
H=x^2y^2+z^2x^2=x^2(y^2+z^2)>=2(x^2)yz
要证(x^2y^2+y^2z^2+z^2x^2)/(x+y+z)>=xyz
即证(x^2y^2+y^2z^2+z^2x^2)>=(x^2)yz+(y^2)xz+(z^2)xy
又知P+Q+H=2(x^2y^2+y^2z^2+z^2x^2)>=2[(x^2)yz+(y^2)xz+(z^2)xy]
得证.
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