Ta có: $9\ge (a+b+c)^2\ge 3(ab+bc+ca)\implies ab+bc+ca\le 3$. Nên: $VT=\dfrac{1}{a^2+b^2+c^2}+\dfrac{1}{ab+bc+ca}+\dfrac{1}{ab+bc+ca}+\dfrac{2007}{ab+bc+ca} \\ \ge \dfrac{9}{(a+b+c)^2}+\dfrac{2007}{ab+bc+ca}\ge \dfrac{9}{3^2}+\dfrac{2007}{3}=670. \blacksquare$