设a,b,c均为正数,且a+b+c=1,证明:1/a+1/b+1/c≥9.

来源:学生作业帮助网 编辑:作业帮 时间:2024/04/28 01:46:43
设a,b,c均为正数,且a+b+c=1,证明:1/a+1/b+1/c≥9.

设a,b,c均为正数,且a+b+c=1,证明:1/a+1/b+1/c≥9.
设a,b,c均为正数,且a+b+c=1,证明:1/a+1/b+1/c≥9.

设a,b,c均为正数,且a+b+c=1,证明:1/a+1/b+1/c≥9.
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

a+b+c=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*√[(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

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