a,b,c＞0则x=a+1/b,y=b+1/c,z=c+1/a,求x,y,z至少有一个大于或等于2

显示全部楼层 · 2012-3-30 15:06:01

x=a+1/b,y=b+1/c,z=c+1/a三式相加x+y+z=a+b+c+1/a+1/b+1/c>=6及x/3+y/3+z/3>=2
(1)若x,y,z均小于2，则x/3+y/3+z/3<2,
(2)(2)与（1）矛盾，所以x,y,z至少有一个大于或等于2 【祝开开心心，万事如意！O(∩_∩)O】可以追问，请及时采纳！(*^__^*) 谢谢！...

千问 · 2012-3-30 15:06:01

x+y+z=a+1/b+b+1/c+c+1/a
=a+b+c+(1/b+1/c+1/a)
=a+b+c+(ac+ab+bc)/abc
≥33√abc+33√a2b2c2/abc
=33√abc+33√1/abc...