已知x,y,z为正实数,求3（x^2+y^2+z^2）+2/x+y+z的最小值.好像要用柯西不等式做.

题目

已知x,y,z为正实数,求3（x^2+y^2+z^2）+2/x+y+z的最小值.好像要用柯西不等式做.

答案

若x、y、z∈R+,则
依三元均值不等式和柯西不等式得：
3(x²+y²+z²)+2/(x+y+z)
＝(1²+1²+1²)(x²+y²+z²)+2/(x+y+z)
≥(x+y+z)²+1/(x+y+z)+1/(x+y+z)
≥3·[(x+y+z)²·1/(x+y+z)·1/(x+y+z)]^(1/3)
＝3.
以上两不等号同时取等时,易得x＝y＝z＝1/3.
∴当x＝y＝z＝1/3时,所求最小值为：3.