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!

Solution
$$\frac{4x^2+8x+13}{6(x+1)}=\frac{4(x+1)^2+9}{6(x+1)}=\frac{2}{3}(x+1)+\frac{3}{2(x+1)}\geq 2$$
with the equality holding if and only if $\dfrac{2}{3}(x+1)=\dfrac{3}{2(x+1)}$,namely, $x=\dfrac{1}{2}.$
