a>0.b>0,c>0 a+b+c=1,请证明1/(a+b)+1/(b+c)+1/(a+c)≥9/2

a>0.b>0,c>0 a+b+c=1,请证明1/(a+b)+1/(b+c)+1/(a+c)≥9/2
a>0.b>0,c>0 a+b+c=1,请证明1/(a+b)+1/(b+c)+1/(a+c)≥9/2

a>0.b>0,c>0 a+b+c=1,请证明1/(a+b)+1/(b+c)+1/(a+c)≥9/2
解,法一：因为2(a+b+c)=2,所以由柯西不等式
[(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