一道数学难题a,b,c均为正数,且a+b+c=1,求(a+1/a)2+(b+1/b)2+(c+1/c)2的最小值.[(a+1/a)2表示(a+1/a)的平方].
2019-05-22
一道数学难题
a,b,c均为正数,且a+b+c=1,求(a+1/a)2+(b+1/b)2+(c+1/c)2的最小值.[(a+1/a)2表示(a+1/a)的平方].
优质解答
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
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