分类：教育/科学-学习帮助
问题：分式不等式设x、y、z∈R+,且x+y+z＝1.证明：x/(x+ 
设x、y、z∈R+,且x+y+z＝1.
证明：x/(x+yz)+y/(y+zx)+z/(z+xy)≤9/4.
回答：证明：∑x/(x+yz)≤9/4
<＝＝>∑[x(x+y+z)/(x(x+y+z)+yz)]≤9/4
<＝＝>∑[x(x+y+z)/(x+y)(x+z)]≤9/4
<＝＝>(x+y+z)∑x(y+z)≤9(x+y)(y+z)(z+x)/4
<＝＝>8(x+y+z)(xy+yz+zx)≤9(x+y)(y+z)(z+x)
<＝＝>∑(x(y-z)^2≥0.
此式显然成立，故原不等式成立。