题目内容：

设a,b,c均为正实数,求证：a/(b+c)+b/(a+c)+c/(a+b)大于等于3/2

优质解答

证明：a/(b+c)+b/(a+c)+c/(a+b)=(a+b+c)/(b+c)-1+(a+b+c)/(b+c)-1+(a+b+c)/(a+c)-1=(a+b+c)(1/(b+c)+1/(a+c)+1/(a+b))-3=1/2((a+b)+(b+c)+(a+c))(1/(b+c)+1/(a+c)+1/(a+b))-3
根据柯西不等式,((a+b)+(b+c)+(a+c))(1/(b+c)+1/(a+c)+1/(a+b))>=9
所以原式>=3/2