分类：教育/科学-理工学科-数学
问题：不等式正实数a、b、c满足a+b+c=3,证明(a+1/a)^3 
正实数a、b、c满足a+b+c=3,证明
(a+1/a)^3+(b+1/b)^3+(c+1/c)^3≥24.
回答：证明:

依Cauchy不等式得
[(a+1/a)+(b+1/b)+(c+1/c)][(a+1/a)^3+(b+1/b)^3+(c+1/c)^3]
≥[(a+1/a)^2+(b+1/b)^2+(c+1/c)^2]^2
≥{1/3[(a+1/a)+(b+1/b)+(c+1/c)]^2}^2
∴(a+1/a)^3+(b+1/b)^3+(c+1/c)^3
≥1/9[(a+1/a)+(b+1/b)+(c+1/c)]^3
=1/9[(a+b+c)+1/3(a+b+c)(1/a+1/b+1/c)]^3
=1/9{3+1/3[3+(a/b+b/a)+(b/c+c/b)+(c/a+a/c)]}^3
≥1/9{3+1/3[3+2+2+2]}^3
=216/9
=24.
证毕.


其实,这是米特里诺维奇-乔科维奇不等式的特例,难度不大,有好几种证法.