正数a,b,c满足a+b+c=1,求证 (a+1/a)(b+1/b)(c+1/c)>=1000/27

题目

正数a,b,c满足a+b+c=1,求证 (a+1/a)(b+1/b)(c+1/c)>=1000/27

答案

证明：a+1/a=a+1/(9a)+1/(9a)+...+1/(9a)(9个1/9a相加)≥10*((1/9)^9/a^8)^(1/10)
同理b+1/b≥10*((1/9)^9/b^8)^(1/10)
c+1/c≥10*((1/9)^9/c^8)^(1/10)
以上三式相乘,∵1=a+b+c>=3(abc)^(1/3),∴1/(abc)>=3^3.
(a+1/a)(b+1/b)(c+1/c)≥1000*((1/9)^27/(abc)^8)^(1/10)≥
1000*((1/9)^27*3^24)^(1/10)≥1000*((1/3)^30)^(1/10)≥1000/27