Question

a,b,c为实数，证明a/（b+c)+b/(a+c)+c/(a+b)>=3/2

Answer 1

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
证毕
或利用柯西不等式
[c/(a+b)+a/(b+c)+b/(a+c)]*[c(a+b)+a(b+c)+b(a+c)]>=(a+b+c)^2
而[c(a+b)+a(b+c)+b(a+c)]=2(ab+bc+ac)<=2/3*(a+b+c)^2
这是因为(a-b)^2+(b-c)^2+(a-c)^2>=0
所以c/(a+b)+a/(b+c)+b/(a+c)>=(a+b+c)^2/[2/3*(a+b+c)^2]=3/2

Answer 2

a,b,c是正实数,用柯西不等式最简:[(a+b)+(b+c)+(c+a)]*[1/(a+b)+1/(b+c)+1/(c+a)]>=(1+1+1)^2=9<==>(a+b+c)[1/(a+b)+1/(b+c)+1/(c+a)]>=9/2<==>(a+b+c)/(a+b)+(a+b+c)/(b+c)+(a+b+c)/(c+a)>=9/2<==>c/(a+b)+b/(c+a)+a/(b+c)>=3/2