Question

Cho a,b,c>0. Tìm min:

\(A=\dfrac{a+1008}{b+c+2016}+\dfrac{b+1008}{c+a+2016}+\dfrac{c+1008}{a+b+2016}\)

Answer 1

\(đặt:a+1008=x;b+1008=y;c+1008=z\)

\(\Rightarrow A=\dfrac{x}{y+z}+\dfrac{y}{x+z}+\dfrac{z}{x+y}=\dfrac{x^2}{xy+xz}+\dfrac{y^2}{xy+yz}+\dfrac{z^2}{xz+yz}\ge\dfrac{\left(x+y+z\right)^2}{2\left(xy+yz+xz\right)}\ge\dfrac{\left(x+y+z\right)^2}{2.\dfrac{\left(x+y+z\right)^2}{3}}=\dfrac{3}{2}\)

\(dấu"="\Leftrightarrow x=y=z\Leftrightarrow a=b=c\)