Smallest Value!

Let $\ u = 1+x$ $\frac{4x^2 +8x+13}{6(1+x)} = \frac{4u^2 +9}{6u}$ Now $\ (2u-3)^2 \geq\ 0 \implies\ 4u^2 +9 \geq\ 12u$ . Dividing the inequality by 6u (note that the inequality sign can't reverse as $\ u \geq\ 1$ ) gives: $\frac{4u^2 +9}{6u} \geq\ 2$ $\therefore\ min \ value = 2$

$\begin{aligned} \frac{4x^2+8x+13}{6(1+x)} & = \frac{4(1+x)^2 + 9}{6(1+x)}\\ & = \frac{2(1+x)}{3} + \frac{3}{2(1+x)} \quad \quad \small \color{#3D99F6}{\text{Let } y = \frac{2(1+x)}{3}} \\ & = y + \frac{1}{y} \quad \quad \quad \quad \quad \quad \quad \quad \space \space \small \color{#3D99F6}{\text{Since } y > 0 \text{, we can apply AM-GM inequality.}} \\ \Rightarrow \frac{4x^2+8x+13}{6(1+x)} & \ge \boxed{2} \end{aligned}$

The fraction can be written as A = (4(x^2 + 2x + 1) + 9 ) / (6(1+x)) = 2(x+1)/3 + 3/2(x+1). Now use the AM - GM inequality : We have A is greater or equal to 2