Question

已知a+b+c=1且abc都为正数。求（a+1/a)2+(b+1/b)2+(c+1/c)2的最小值

Answer 1

因a²+1/81a²≥2/9,b²+1/81b²≥2/9,c²+1/81c²≥2/9
则a²+b²+c²+(1/a²+1/b²+1/c²)/81≥2/3
则（a+1/a)²+(b+1/b)²+(c+1/c)²
=6+a²+1/a²+b²+1/b²+c²+1/c²
≥6+2/3+80(1/a²+1/b²+1/c²)/81
又1/a²+1/b²+1/c²≥3(1/abc)^(2/3)
又1=a+b+c≥3(abc)^(1/3)
=>(1/abc)^(2/3)≥9
则1/a²+1/b²+1/c²≥27
则原式≥20/3+80*27/81=100/3

Answer 2

a=b=c=1/3时最小
得出（a+1/a)^2+(b+1/b)^2+(c+1/c)^2的最小值
=（1/3+3)^2+(1/3+3)^2+(1/3+3)^2
=3*100/9
=100/3

Answer 3

a=b=c=1/3时最小
得出（a+1/a)2+(b+1/b)2+(c+1/c)2的最小值为24

Answer 4

a+1/a)2+(b+1/b)2+(c+1/c)2=2(a+b+c)+2(1/a+1/b+1/c)
=2+2(a+b+c)(1/a+1/b+1/c)=2+2(1+a/b+a/c+
b/a+1+b/c+1/a+1/b+1/c)≥
2+2(3+3均直定理)=20

Answer 5

好象是20吧!我用的口算,不一定准!

Answer 6

因a²+1/81a²≥2/9,b²+1/81b²≥2/9,c²+1/81c²≥2/9
则a²+b²+c²+(1/a²+1/b²+1/c²)/81≥2/3
则（a+1/a)²+(b+1/b)²+(c+1/c)²
=6+a²+1/a²+b²+1/b²+c²+1/c²
≥6+2/3+80(1/a²+1/b²+1/c²)/81
又1/a²+1/b²+1/c²≥3(1/abc)^(2/3)
又1=a+b+c≥3(abc)^(1/3)
=>(1/abc)^(2/3)≥9
则1/a²+1/b²+1/c²≥27
则原式≥20/3+80*27/81=100/3