a+b=c=1证明根号A＋根号B＋根号C＜＝根号3

证明：利用算术平均<=平方平均(An<=Qn)得：
根号a+根号b+根号c<=3根号[(a+b+c)/3]=根号3

另证：利用Cauchy不等式内积<=模长之积，得
1*根号a+1*根号b+1*根号c<=根号(1^2+1^2+1^2)*根号(a+b+c)=根号3

题目错了吧 a+b+c=1

柯西不等式(∑ai^2) * (∑bi^2) ≥ (∑ai *bi)^2.

（1+1+1）*（a+b+c）≥（根号A＋根号B＋根号C）²

=》根号A＋根号B＋根号C＜＝根号3

因为:(根号a＋根号b＋根号c)^2
=a+b+c+2根号ab+2根号ac+2根号bc
=1+2根号ab+2根号ac+2根号bc
＜＝1+(a+b)+(a+c)+(b+c)
=3
所以:根号a＋根号b＋根号c＜＝根号3