题目

X y z >0.Yz+zx+xy=1.求证：x(1-y^2)(1-z^2)+y(1-z^)(1-x^2)+z(1-x^2)(1-y^2)

提问时间：2020-10-12

答案

x(1-y^2)(1-z^2)+y(1-z^2)(1-x^2)+z(1-x^2)(1-y^2)
=(x +x y^2 z^2 - xy^2 - x z^2)+(y +y x^2 z^2 - y x^2- y z^2 )+(z +z x^2 y^2 -z y^2 - z x^2 )
=x(1-xy -xz)+ y(1- xy -yz)+z(1-xz-yz) + xyz (xy +yz +xz)
∵yz+zx+xy=1
∴上式=x(yz+zx+xy-xy -xz)+ y(yz+zx+xy- xy -yz)+z(yz+zx+xy-xz-yz) + xyz (xy +yz +xz)
=xyz +xyz + xyz +xyz
=4xyz
=4(xy*yz*xz)^1/2)
≤4×(（yz+zx+xy/3）^3)^1/2
=4×(（1/3）^3)^1/2
=4/9倍根号3