Question

已知a>0b>0c>0且a+b+c=1求证1/a+b+1/b+c+1/c+a>=9/2

Answer 1

法一：因为2(a+b+c)=2,所以由Cauchy不等式
[(a+b)+(b+c)+(c+a)][1/(a+b)+1/(b+c)+1/(a+c)]>=
(1+1+1))^2=9
即2[1/(a+b)+1/(b+c)+1/(a+c)]>=9
所以1/(a+b)+1/(b+c)+1/(a+c)>=9/2
法二：
把a+b+c=1代入1/(a+b)+1/(b+c)+1/(a+c)>=9/2
得2a/(b+c)+2b/(a+c)+2c/(a+b)>=3
由对称性不妨设a<=b<=c,则a+b<=a+c<=b+c,1/(b+c)<=1/(a+c)<=1/(a+b),由排序不等式正序和>=乱序和>=逆序和，有
a/(b+c)+b/(a+c)+c/(a+b)>=b/(b+c)+c/(a+c)+a/(a+b)
a/(b+c)+b/(a+c)+c/(a+b)>=c/(b+c)+a/(a+c)+b/(a+b)
两式相加得2a/(b+c)+2b/(a+c)+2c/(a+b)>=3
所以原不等式成立。证毕！