Inequalities 2

$\frac{4x²+4x+4x+4+9}{6(1+x)}$ = $\frac{4x(x+1)}{6(x+1)}$ + \frac{4(x+1)}{6(x+1)}+\(\frac{9}{6(x+1)}

So,

$\frac{2x}{3}$ + $\frac{2}{3}$ + $\frac{3}{2(x+1)}$ = $\frac{2(x+1)}{3}$ + $\frac{3}{2(x+1)}$

Now we have two terms and we need to apply here AM≥GM

Thus by applying this inequality we get the answer as $\boxed{2}$ .

We have:

$\dfrac{4x^2+8x+13}{6+6x}=2+\dfrac{(2x-1)^2}{6x+6}\ge2, \forall x\ge0$ .

The equality holds when $x=\dfrac{1}{2}$ .

So, the answer is $\boxed{2}$