Question

已知a、b、c为互不相等的正数，且a+b+c=1，求证1/a+1/b+1/c＞9

Answer 1

把分子的1替换为a+b+c
1/a+1/b+1/c
=(a+b+c)/a+(a+b+c)/b+(a+b+c)/c
=1+b/a+c/a+a/b+1+c/b+a/c+b/c+1
=3+(b/a+a/b)+(a/c+c/a)+(b/c+c/b)
>=3+2+2+2=9
但a,b,c不等，所以等号取不到

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