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!

$$\frac {4x^2+8x+13}{6(x+1)}$$
$$= \frac {2(x+1)}{3} + \frac {3}{2(x+1)}$$
$$ = y+\frac {1}{y} \ge 2$$
