设a,b,c都大于0 1.求证：c/a+a/(b+c)+b/c≥2 2.求4/a+1/b+1/c+（a+b+c)^2的最小值

题目

设a,b,c都大于0 1.求证：c/a+a/(b+c)+b/c≥2 2.求4/a+1/b+1/c+（a+b+c)^2的最小值
运用柯西不等式解答

答案

1)
c/a+a/(b+c)+b/c+1-1
=c/a+a/(b+c)+(b+c)/c-1
利用均值不等式x+y+z>=3(xyz)^(1/3)
=>c/a+a/(b+c)+(b+c)/c-1>=3(c/a*a/(b+c)*(b+c)/c)^(1/3)-1
=3-1=2
当且仅当c/a=a/(b+c)=(b+c)/c时等号成立