条件：正实数满足a+b+c=1①.求(a+1/a)²+(b+1/b)²+(c+1/c)²的最小值②.若a²/(1+a)+b²/(1+b)+c²/(1+c)=1/4 求abc的值
问题描述：
条件：正实数满足a+b+c=1
①.求(a+1/a)²+(b+1/b)²+(c+1/c)²的最小值
②.若a²/(1+a)+b²/(1+b)+c²/(1+c)=1/4 求abc的值
答
由柯西不等式
(a+b+c)(1/a+1/b+1/c)>=(a*1/a+b*1/b+c*1/c)^2=(1+1+1)^2=9
a+b+c=1
所以1/a+1/b+1/c>=9
又由柯西不等式
[(a+1/a)^2+(b+1/b)^2+(c+1/c)^2](1+1+1)
>=[(a+1/a)*1+(b+1/b)*1+(c+1/c)]^2
=[(a+b+c)+(1/a+1/b+1/c)]^2
=[1+(1/a+1/b+1/c)]^2
>=(1+9)^2=100
即3[(a+1/a)^2+(b+1/b)^2+(c+1/c)^2]>=100
所以(a+1/a)^2+(b+1/b)^2+(c+1/c)^2>=100/3
所以最小值=100/3
答
第一题：【By 西陵楚客】由柯西不等式(a+b+c)(1/a+1/b+1/c)>=(a*1/a+b*1/b+c*1/c)^2=(1+1+1)^2=9a+b+c=1所以1/a+1/b+1/c>=9又由柯西不等式[(a+1/a)^2+(b+1/b)^2+(c+1/c)^2](1+1+1)>=[(a+1/a)*1+(b+1/b)*1+(c+1/c)]^2...