Question

设a,b,c均为正数，且a+b+c=1,证明：1/a+1/b+1/c≥9.

Answer 1

1/a+1/b+1/c
=(a+b+c)/a+(a+b+c)/b+(a+b+c)/c
=3+b/a+c/a+a/b+c/b+a/c+b/c
=3+(b/a+a/b)+(a/c+c/a)+(c/b+b/c)
b/a+a/b>=2*根号[(b/a)*(a/b)]=2
同理
c/a+a/c>=2
c/b+b/c>=2
所以
1/a+1/b+1/c>=9
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Answer 2

a+b+c=1
则
1/a+1/b+1/c
=(a+b+c)/a+(a+b+c)/b+(a+b+c)/c
=3+b/a+c/a+a/b+c/b+a/c+b/c
=3+(b/a+a/b)+(a/c+c/a)+(c/b+b/c)
b/a+a/b≥2*√[(b/a)*(a/b)]=2
同理
c/a+a/c≥2
c/b+b/c≥2
所以
1/a+1/b+1/c≥9

Answer 3

原式=（a+b+c）/a+（a+b+C）/b+（a+b+c）/C=3+（b/a+a/b）+（b/c+c/b）+（c/a+a/c）≥3+2+2+2=9