分类：教育/科学-学习帮助
问题：94香港奥数题设x、y、z≥0,且满足xy+yz+zx＝1.求证 
设x、y、z≥0,且满足xy+yz+zx＝1.求证：
x(1-y^2)(1-z^2)+y(1-z^2)(1-x^2)+z(1-x^2)(1-y^2)≤4√3/9.
回答：证明：
展开不等式左边 
x(1-y^2)(1-z^2)+y(1-z^2)(1-x^2)+z(1-x^2)(1-y^2) 
= (x+y+z)+(xy^2z^2+x^2yz^2+x^2y^2z)-(xy^2+xz^2+yz^2+yx^2+zx^2+zy^2)
= (x+y+z)+(xy+zy+xz)xyz-(xy^2+xz^2+yz^2+yx^2+zx^2+zy^2)
= (x+y+z)+xyz-(xy^2+xz^2+yz^2+yx^2+zx^2+zy^2)..........⑴

又 xy+yz+zx＝1
两边同时乘以x+y+z有 (x+y+z)(xy+yz+zx) = x+y+z 
即 3xyz+(xy^2+xz^2+yz^2+yx^2+zx^2+zy^2) = x+y+z
所以 xy^2+xz^2+yz^2+yx^2+zx^2+zy^2 = (x+y+z)-3xyz......⑵

⑵代入⑴中有不等式左边 
x(1-y^2)(1-z^2)+y(1-z^2)(1-x^2)+z(1-x^2)(1-y^2) 
= (x+y+z)+xyz-[(x+y+z)-3xyz]
= 4xyz................................................⑶


又 xy+yz+zx＝1
由平均值不等式有 1 = xy+yz+zx ≥　3[(xy)(yz)(zx)]^(1/3) 
即得 xyz ≤ √3/9......................................⑷

将⑷代入⑶中，有不等式左边 
x(1-y^2)(1-z^2)+y(1-z^2)(1-x^2)+z(1-x^2)(1-y^2) 
= 4xyz
≤ 4√3/9

证毕