【證法1】左邊=c/(a+b)+1+a/(b+c)+1+b/(c+a)+1-3
=(a+b+c)/(a+b)+(a+b+c)/(b+c)+(a+b+c)/(c+a)-3
=(a+b+c)[1/(a+b)+1/(b+c)+1/(c+a)]-3
=1/2*[(a+b)+(b+c)+(c+a)][1/(a+b)+1/(b+c)+1/(c+a)]-3
由柯西不等式
≥1/2*[(a+b)*1/(a+b)+(b+c)*1/(b+c)+(c+a)*1/(c+a)]^2-3
=3/2
【證法2】c/(a+b)+b/(c+a)+a/(b+c)
=c/(a+b)+b/(c+a)+a/(b+c)+[(a+b)/(a+b)+(c+a)/(c+a)+(b+c)/(b+c)]-3
=(a+b+c)/(a+b)+(a+b+c)/(b+c)+(a+b+c)/(c+a)-3
=(a+b+c))*[1/(a+b)+1/(b+c)+1/(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
所以c/(a+b)+a/(b+c)+b/(c+a)>=3/2