smc

To begin, lets denote the equation, $\frac{4x^2+8x+13}{6(1+x)}$ as $f(x)$. Let's notice that: $$\begin{array}{rl}f(x)& =\frac{4x^2+8x+13}{6(1+x)}\\ & \\ & =\frac{4(x^2+2x)+13}{6(x+1)}\\ & \\ & =\frac{4(x^2+2x+1-1)+13}{6(x+1)}\\ & \\ & =\frac{4(x+1{)}^2+9}{6(x+1)}\\ & \\ & =\frac{4(x+1{)}^2}{6(x+1)}+\frac{9}{6(1+x)}\\ & \\ & =\frac{2(x+1)}{3}+\frac{3}{2(x+1)}\\ & \end{array}$$ After this simplification, we may notice that we may use calculus, or the AM-GM inequality to finish this problem because $x\ge 0$, which implies that both $\frac{2(x+1)}{3}\text{ and }\frac{3}{2(x+1)}$ are greater than zero. Continuing with AM-GM: $$\begin{array}{rl}\frac{\frac{2(x+1)}{3}+\frac{3}{2(x+1)}}{2}& \ge \sqrt{\frac{2(x+1)}{3}\cdot \frac{3}{2(x+1)}}\\ & \\ f(x)& \ge 2\end{array}$$ Therefore, $f(x)=\frac{4x^2+8x+13}{6(1+x)}\ge 2$, $\boxed{2}$ $(\text{With equality when }\frac{2(x+1)}{3}=\frac{3}{2(x+1)},\text{ or }x=\frac{1}{2})$