根据数值的对称性,可令a≥b≥c,
则有1/(b+c)≥1/(a+c)≥1/(a+b)
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)
两式相加可得2[a/(b+c)+b/(a+c)+c/(a+b)]≥3
所以可求得a/(b+c)+b/(a+c)+c/(a+b)≥3
左边=(a+b+c)/(a+b)+(a+b+c)/(b+c)+(a+b+c)/(c+a)-3=0.5*(a+b+b+c+c+a)*[1/(a+b)+1/(b+c)+1/(c+a)]-3≥0.5*{3*[(a+b)(b+c)(c+a)]^1/3}*{3*[1/(a+b)*1/(b+c)*1/(c+a)]^1/3}-3=0.5*3*3-3=3/2
你上学时数学就好吧