Đặt $x=\frac{1}{a},y=\frac{1}{b},z=\frac{1}{c}$.Từ giả thiết có:$9(x^2+y^2+z^2)=3(xy+yz+xz)+2014\leq 3(x^2+y^2+z^2)+2014\Rightarrow \frac{1007}{3}\geq x^2+y^2+z^2$
Do vậy $(x+y+z)^2=(x^2+y^2+z^2)+2(xy+yz+xz)\leq 3(x^2+y^2+z^2)\leq 1007$
$5a^2+2ab+2b^2=4a^2+2ab+b^2+(a^2+b^2)\geq 4a^2+4ab+b^2=(2a+b)^2$
$\Rightarrow \frac{1}{\sqrt{5a^2+2ab+2b^2}}\leq \frac{1}{2a+b}\leq \frac{1}{9}(\frac{2}{a}+\frac{1}{b})=\frac{2x+y}{9}$
Tương tự ta có:$VT\leq \frac{x+y+z}{3}\leq \frac{1}{3}.\sqrt{1007}$
Dấu = xảy ra khi $x=y=z$