∵a+b+c=1
∴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
根據(jù)基本不等式:
∵b/a+a/b≥2
c/a+a/c≥2
c/b+b/c≥2
∴3+b/c+c/b+a/c+c/a+a/b+b/a
≥3+2+2+2=9
當且僅當a=b=b時成立
∵a、b、c為為互不相等的正實數(shù)
∴＞9