Question

a,b,c∈R+ 求证c/(a+b)+a/(b+c)+b/(a+c)≥3/2

Answer 1

证明：c/(a+b)+a/(b+c)+b/(a+c) = (a+b+c)(1/(a+b) + 1/(b+c) + 1/(a+c)) - 3
=1/2 * (a+b + b+c + c+a)(1/(a+b) + 1/(b+c) + 1/(a+c)) - 3
(柯西不等式) ≥1/2 * (1 + 1 + 1)^2 -3 = 3/2

Answer 2

c/(a+b)+a/(b+c)+b/(a+c)≥3[abc/(a+b)(b+c)(a+c)]^(1/3)，此时c/(a+b)=a/(b+c)=b/(a+c)=(a+b+c)/2(a+b+c)=1/2，则c/(a+b)+a/(b+c)+b/(a+c)≥3[abc/(a+b)(b+c)(a+c)]^(1/3)=3[(1/2)³]^(1/3)=3/2，c/(a+b)+a/(b+c)+b/(a+c)≥3/2成立。