数学不等式证明.已知x+y+z=1,求证：x^2/[y+2z]+y^2/[z+2x]+z^2/[x+2y]不小于1/3.

运用柯西不等式证明
因为x+y+z=1,
所以x^2/[y+2z]+y^2/[z+2x]+z^2/[x+2y]
=3(x+y+z){x^2/[y+2z]+y^2/[z+2x]+z^2/[x+2y]}/3
=(y+2z+z+2x+x+2y){x^2/[y+2z]+y^2/[z+2x]+z^2/[x+2y]}/3
>=(x+y+z)^2/3
=1/3
原不等式得证