应该是f（a,b,c）=1/(a+b)+1/(b+c)+1/(a+c)>=（3/2）根号3
f（a,b,c）=1/(a+b)+1/(b+c)+1/(a+c)=(ab+bc+ca）[1/(a+b)+1/(b+c)+1/(a+c)]=
a+b+c+[ab/(a+b)]++[ac/(a+c)]++[bc/(b+c)]
先证明a+b+c>=根号3,（a+b+c）^2=a^2+b^2+c^2+2ab+2bc+2ac≥3（ab+bc+ca）=3,所以a+b+c>=根号3.同理可证[ab/(a+b)]>=(根号3)/6,[ac/(a+c)]>=(根号3)/6,
[bc/(b+c)]>=(根号3)/6,上述等号成立的前提是a=b=c=(根号3)/3,
综上所述,f（a,b,c）=1/(a+b)+1/(b+c)+1/(a+c)>=（3/2）根号3
即f（a,b,c）=1/(a+b)+1/(b+c)+1/(a+c)>5/2,其中5/2不可取.