a+b+c=1,求证(1+1/a)(1+1/b)(1+1/c)>=64

(1+1/a)(1+1/b)(1+1/c)
=1+ (1/a+1/b+1/c) + (1/ab+1/bc+1/ca) +1/abc
=1+ (1/a+1/b+1/c) + (a+b+c)/abc +1/abc
=1+ (1/a+1/b+1/c) + 2/abc

其中由柯西不等式，
(1/a+1/b+1/c)(a+b+c) > =(1+1+1)^2 = 9，
而a+b+c=1，所以(1/a+1/b+1/c) >= 9。

由几何不等式，
a+b+c=1 >= 3(abc)^1/3，
所以abc <= 1/27 1/abc >= 27，

因此
(1+1/a)(1+1/b)(1+1/c)=1+ (1/a+1/b+1/c) + 2/abc >= 1+9+2*27=64。

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