a,b,c为正数,a+b+c=1证:(1) (1/a+1)(1/b+1)(1/c+1)>=64 (2)(1/a-1)(1/b-1)(1/c-1)>=8

1)a+b+c=1≥3(abc)^1/3

abc≤1/27 1/abc≥27

(1/a+1)(1/b+1)(1/c+1)

=1/a+1/b+1/c+1/ab+1/bc+1/ac+1+1/abc≥3(1/abc)^1/3+3
(1/abc)^2/3+1/abc+1=64

所以(1/a+1)(1/b+1)(1/c+1)≥64得证

2)(1/a-1)
=(1-a)/a
=(a+b+c-a)/a
=(b+c)/a
又(√b-√c)^2≥0
b+c≥2√(bc)
∴（1/a-1)=(b+c)/a≥2√(bc)/a
同理
(1/b-1)≥2√(ac)/b
(1/c-1)≥2√(ab)/c

故(1/a-1)*(1/b-1)*(1/c-1)≥[2√(bc)/a]*[2√(ac)/b]*[2√(ab)/c]
=8 √[(a^2)*(b^2)8(c^2)]/(abc)
=8

∴(1/a-1)*(1/b-1)*(1/c-1)≥8