已知a,b,c属于R,a+b+c=1,求证:1/(a+1)+1/(b+1)+(c+1)>=9/4

已知a,b,c属于R,a+b+c=1,求证:1/(a+1)+1/(b+1)+(c+1)>=9/4
已知a,b,c属于R,a+b+c=1,求证:1/(a+1)+1/(b+1)+(c+1)>=9/4

已知a,b,c属于R,a+b+c=1,求证:1/(a+1)+1/(b+1)+(c+1)>=9/4
两边同乘4左边写成：[(a+1)+(b+1)+(c+1)]*[1/(a+1)+1/(b+1)+1/(c+1)]展开得3+(a+1)/(b+1)+(b+1)/(a+1)+（a+1）/(c+1）+（c+1）/(a+1)+(c+1)/(b+1)+(b+1)/(c+1)>=9(用均值不等试）推广为（a1+a2+a3.+an)*[(1/(a1)+1/(a2)+1/(a3)+...1/(an)]>=n*n(各数为正数）