Question

已知x+y+z=1,x,y,z大于0,求证:x^2/(y+z)+y^2/(z+x)+z^2/(y+x)大于等于1/2

Answer 1

解：利用柯西不等式
∵x,y,z大于0，x+y+z=1
∴x^2/(y+z)+y^2/(z+x)+z^2/(y+x)
=｛[x/√（y+z)]²+ [ y/√(z+x)]²+[z/√(y+x)]²｝[（√（y+z））²+（√（z+x））²+（√（y+x））²]≥（x+y+z)²
∴x^2/(y+z)+y^2/(z+x)+z^2/(y+x)
≥（x+y+z)²/2（x+y+z)=1/2
当且仅当x=y=z=1/3时等号成立

Answer 2

利用柯西不等式
[x^2/(y+z)+y^2/(x+z)+z^2/(y+x)][(y+z)+(x+z)+(x+y)}>=(√[x^2/(y+z)*(y+z)]+√[y^2/(y+z)*(y+z)]+√[z^2/(x+y)*(x+y)])^2=(x+y+z)^2
x^2/(y+z)+y^2/(x+z)+z^2/(x+y)>=(x+y+z)^2/(2x+2y+2z)=1/2
等号当且仅当x=y=z=1/3时成立