Question

已知a＞0，b＞0，c＞0。求证a/bc+b/ac+c/ab≥1/a+1/b+1/c

貌似要运用均值不等式

Answer 1

a/bc+b/ac+c/ab
=(a²/abc+b²/abc+c²）/abc
=(a²+b²+c²)/abc

1/a+1/b+1/c
=bc/abc+ac/abc+ab/abc
=(bc+ac+ab)/abc

a²+b²+c²-(bc+ac+ab)
=（2a²+2b²+2c²-2bc-2ac-2ab）/2
=[（a-b)²+(a-c)²+(b-c)²]/2≥0

所以a²+b²+c²≥bc+ac+ab
所以a/bc+b/ac+c/ab≥1/a+1/b+1/c

Answer 2

根据对称性设a>b>c
所以ab>ac>bc
因为排序不等式得
a/bc+b/ac+c/ab>a/ab+b/bc+c/ac=1/a+1/b+1/c
所以原式得证
当a=b=c时等号成立