若x、y、z∈R+,則
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.
以上兩不等號同時取等時,易得x＝y(tǒng)＝z＝1/3.
∴當(dāng)x＝y(tǒng)＝z＝1/3時,所求最小值為：3.