Question

设a,b,c为正数,求证:1/a+1/b+1/c>=9/(a+b+c)

Answer 1

(a+b+c)(1/a+1/b+1/c)
=(a+b+c)/a+(a+b+c)/b+(a+b+c)/c
=3+(a/b+b/a)+(a/c+c/a)+(b/c+c/b)
由
均值不等式
a/b+b/a>=2根号(a/b*b/a)=2
同理a/c+c/a>=2
b/c+c/b>=2
所以原式>=3+2+2+2
当且仅当a=b=c时等号成立
所以(a+b+c)(1/a+1/b+1/c)>=9
所以1/a+1/b+1/c>=9/(a+b+c)

Answer 2

已知a,b,c属于正实数,且a+b+c=1,求证1/a+1/b+1/c大于等于9
1/a+1/b+1/c
=(a+b+c)/a+(a+b+c)/b+(a+b+c)/c
=1+(b+c)/a+1(a+c)/b+1(a+b)/c
=3+b/c+c/b+a/c+c/a+a/b+b/a
（由于b/a+a/b>=2，c/a+a/c>=2，c/b+b/c>=2）
>=3+2+2+2
=9