với a,b,c>0
áp dung bđt \(\frac{a^2}{x}+\frac{b^2}{y}\ge\frac{\left(a+b\right)^2}{x+y}\)( bđt svacxo) ta có :
A=\(\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\ge\frac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}\)\(=\frac{a+b+c}{2}=\frac{2016}{2}=1008\)
=> min A=1008 dấu bằng xảy ra <=>a=b=c=672