Question

设a,b,c均为正数,且a+b+c=1.证明:1/a+1/b+1/c≥9

过程要详细哦！我刚开始学不等式。谢谢 ！！！！！！！！

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

Answer 2

因为a+b+c=1将二式中1换成a+b+c得(b/a+a/b)+(c/b+b/c)+(a/c+c/a)+3因为b/a+a/b≥2，当且仅当a=b时取=号同理(b/a+a/b)+(c/b+b/c)+(a/c+c/a)≥6则(b/a+a/b)+(c/b+b/c)+(a/c+c/a)+3≥9即所证成立