Question

已知实数x，y，z满足x2＋y2＋z2＝2求证，x＋y＋z≤xyz＋2？

竞赛题有难度，注意是x²＋y²＋z²＝2

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】

Answer 2

由2x+2y+2z＝2=2*(x+y+z)，得x+y+z=1，当xyz都为实数时，xyz＞0，得xyz+2＞2，因为2＞1，所以x+y+z小于等于(小于等于等同于不大于)，所以x+y+z不大于xyz+2