- 不等式已知x、y、z∈R+,证明：xy/(x^2+y^2+2z^
- 已知x、y、z∈R+,证明：xy/(x^2+y^2+2z^2)+yz/(2x^2+y^2+z^2)+zx/(x^2+2y^2+z^2)≤3/4.
- 证明：
由均值不等式得
xy/(x^2+y^2+2z^2)+yz/(2x^2+y^2+z^2)+zx/(x^2+2y^2+z)
=xy(x^2+y^2+z^2+z^2)+yz/(x^2+x^2+y^2+z^2)+zx/(x^2+y^2+y^2+z^2)
≤xy/2根[(y^2+z^2)(z^2+x^2)]+yz/2根[(z^2+x^2)(x^2+y^2)]+zx/2根[(x^2+y^2)(z^2+x^2)]
≤1/4*{[x^2/(z^2+x^2)+y^2/(y^2+z^2)]+[z^2/(z^2+x^2)+y^2/(x^2+y^2)]+[x^2/(x^2+y^2)+z^2/(y^2+z^2)]}
=3/4.
证毕.