Question

已知a,b,c为正实数，求证(a+b+c)(1/a+1/b+1/c)大于等于9，没有a+b+c=1这一条件

Answer 1

1.已知a,b,c属于正实数,求证bc/a+ac/b+ab/c大于等于a+b+c.
bc/a+ac/b+ab/c≥a+b+c.
将上式两边同时乘以2，那么
2bc/a+2ac/b+2ab/c≥2a+2b+2c.
由已知ab≤[（a+b）/2]^2,可以得出a^2+b^2≥2ab
所以bc/a+ac/b≥2c
bc/a+ab/c≥2b
ac/b+ab/c≥2a
将以上三式相加，即可得到结论，所以得证

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

Answer 2

均值定理：
a+b+c≥3³√abc
(1/a+1/b+1/c)≥3³√1/abc
原式≥9³√abc×1/abc=9
得证

Answer 3

(a+b+c)(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
>=3+2+2+2=9

追问

=3+b/c+c/b+a/c+c/a+a/b+b/a
这一步怎么出来哒？

追答

(a+b+c)(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/a+c/a+a/b+c/b+a/c+b/c
=3+b/c+c/b+a/c+c/a+a/b+b/a

追问

3+b/c+c/b+a/c+c/a+a/b+b/a
怎么变成3+2+2+2 ？不理解……

追答

3+b/c+c/b+a/c+c/a+a/b+b/a
>=3+2√(b/c*c/b)+2√(a/c*c/a)+2√(b/a*a/b)
=3+2+2+2
=9