Question

设a,b,c,为正数且a+b+c=1,求证（a+1/a）平方+（b+1/b）平方+（c+1/c)平方大于等于100/3

Answer 1

用柯西不等式

(1+1+1)[(a+1/a)^2+(b+1/b)^2+(c+1/c)^2]>=(a+1/a+b+1/b+c+1/c)^2=(1+1/a+1/b+1/c)^2

(a+b+c)(1/a+1/b+1/c)>=(1+1+1)^2=9

1/a+1/b+1/c>=9

(1+1/a+1/b+1/c)^2>=(1+9)^2=100

(1+1+1)[(a+1/a)^2+(b+1/b)^2+(c+1/c)^2]>=100

(a+a/1)^2+(b+b/1)^2+(c+c/1)^2>=100/3

原式成立