已知正数x,y,z满足x+y+z=1,求证x^2/(y+2z)+y^2/(z+2x)+z^2(x+2y)>=1/3.

Question

已知正数x,y,z满足x+y+z=1,求证x^2/(y+2z) +y^2/(z+2x) +z^2(x+2y)>=1/3.

Answer 1

方法①
根据平均值不等式：
x^2/(y+2z)+(y+2z)/9≥2√{[x^2/(y+2z)][(y+2z)/9]}=2x/3
y^2/(z+2x)+(z+2x)/9≥2√{[y^2/(z+2x)][(z+2x)/9]}=2y/3
z^2/(x+2y)+(x+2y)/9≥2√{[z^2/(x+2y)][(x+2y)/9]}=2z/3
以上3式相加：
x^2/(y+2z)+y^2/(z+2x)+z^2/(x+2y)+3(x+y+z)/9≥2(x+y+z)/3
∵x+y+z=1
∴x^2/(y+2z)+y^2/(z+2x)+z^2/(x+2y)≥1/3
方法②
利用柯西不等式：
[x^2/(y+2z)+y^2/(z+2x)+z^2/(x+2y)][(y+2z)+(z+2x)+(x+2y)]
≥(x+y+z)^2=1
而显然：(y+2z)+(z+2x)+(x+2y)=3(x+y+z)=3
∴x^2/(y+2z)+y^2/(z+2x)+z^2/(x+2y)≥1/3
【以上两方法证明中等号成立的条件都是x=y=z=1/3】