OMC 099(4b) D 因为 \((abc)^{\dfrac 13} \le \dfrac{a+b+c}3\)(基本不等式),将 \(a = xy, b = yz, c = xz\) 代入得到 \((xyz)^{\dfrac 23} \le \dfrac{xy+yz+xz}3 = \langle 2,2,4 \rangle\),所以 \(xyz \le \langle 3,3,6 \rangle\),于是 \(\dfrac1x+\dfrac1y+\dfrac1z = \dfrac{xy+yz…