abc都是正数求证a2/(b+c)+b2/(a+c)+c2/(a+b)>=1/2(a+b+c)

Question

a b c都是正数 求证a2/(b+c)+b2/(a+c)+c2/(a+b)>=1/2(a+b+c)

Answer 1

设a=0.1,b=0.1,c=0.3
a2/(b+c)+b2/(a+c)+c2/(a+b)=0.5
1/2(a+b+c)=1
不成立
若求证a2/(b+c)+b2/(a+c)+c2/(a+b)>=(a+b+c)/2
a2/(b+c)+b2/(a+c)+c2/(a+b)
>a2/(b+c+a)+b2/(a+c+b)+c2/(a+b+c)
=(a2+b2+c2)/(a+b+c)
=(a2/4+b2/4+c2/4+a2/4+b2/4+c2/4+a2/4+b2/4+c2/4+a2/4+b2/4+c2/4)/(a+b+c)
>=(a2/4+b2/4+c2/4+ab/2+bc/2+ac/2+a2/4+b2/4+c2/4)/(a+b+c)
>＝(a+b+c)的平方/2(a+b+c)
=(a+b+c)/2