x,y,z 大于0 且xyz=1 求x2/(y+z)+y2/(x+z)+z2/(x+y)的最小值
问题描述：
x,y,z 大于0 且xyz=1 求x2/(y+z)+y2/(x+z)+z2/(x+y)的最小值
答
利用均值不等式知道x^2/(y+z)+(y+z)/4>=x即x^2/(y+z)>=x-(y+z)/4同理y^2/(x+z)>=y-(x+z)/4z^2/(y+x)>=z-(y+x)/4相加得x2/(y+z)+y2/(x+z)+z2/(x+y)>=x+y+z-(x+y+z)/2=(x+y+z)/2>=3(xyz)^(1/3)/2=3/2当x=y=z时等号成立...