已知a,b,c均为正数且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