正实数x,y,z满足9xyz+xy+yz+zx=4,求证:(1)xy+yz+zx≥4/3;(2)x+y+z≥2.

问题描述:

正实数x,y,z满足9xyz+xy+yz+zx=4,求证:
(1)xy+yz+zx≥

4
3

(2)x+y+z≥2.

证 (1)记t=xy+yz+xz3,∵x,y,z>0.由平均不等式xyz=(3xy•yz•xz)32≤(xy+yz+zx3)32于是4=9xyz+xy+yz+xz≤9t3+3t2,∴(3t-2)(3t2+3t+2)≥0,而3t2+3t+2>0,∴3t-2≥0,即t≥23.∴xy+yz+zx≥43...