Question

数学：帮忙求证不等式.

Answer 1

c/(a+b)+b/(c+a)+a/(b+c)=c/(a+b)+1+b/(c+a)+1+a/(b+c)+1-3=(a+b+c)[1/(a+b)+1/(b+c)+1/(c+a)]-3=[(a+b)+(b+c)+(a+c)][1/(a+b)+1/(b+c)+1/(c+a)]/2-3

>=[1+1+1]^2/2-3 (根据柯西不等式）

=3/2