For $x\geq 0$, what is the smallest value of $\frac{4x^2+8x+13}{6(x+1)}$? I know that I have to use the AM-GM inequality.
I tried separating the fraction: 
$$\frac{4x^2+2x+7}{6(x+1)} + \frac{6(x+1)}{6(x+1)}$$
However, it doesn't seem to make either side of the inequality into a number.
I would appreciate some help, thanks!

Hint: by polynomial euclidean division $\;\dfrac{4x^2+8x+13}{6(x+1)} = \dfrac{1}{6}\left(4(x+1) + \dfrac{9}{x+1}\right) \ge \ldots\,$
