# 1965 AHSME Problems/Problem 34

## Problem 34

For $x \ge 0$ the smallest value of $\frac {4x^2 + 8x + 13}{6(1 + x)}$ is:

$\textbf{(A)}\ 1 \qquad \textbf{(B) }\ 2 \qquad \textbf{(C) }\ \frac {25}{12} \qquad \textbf{(D) }\ \frac{13}{6}\qquad \textbf{(E) }\ \frac{34}{5}$

## Solution 1

To begin, lets denote the equation, $\frac {4x^2 + 8x + 13}{6(1 + x)}$ as $f(x)$. Let's notice that:

\begin{align*} 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{align*}

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{align*} \frac{\frac{2(x+1)}{3} + \frac{3}{2(x+1)}}{2} &\ge {\small \sqrt{\frac{2(x+1)}{3}\cdot \frac{3}{2(x+1)}}}\\\\ f(x) &\ge 2 \end{align*}

Therefore, $f(x) = \frac {4x^2 + 8x + 13}{6(1 + x)} \ge 2$, $\boxed{\textbf{(B)}}$

$(\text{With equality when } \frac{2(x+1)}{3} = \frac{3}{2(x+1)}, \text{ or } x=\frac{1}{2})$

## Solution 2 (Calculus)

Take the derivative of f(x) and f'(x) using the quotient rule. \begin{align*} f(x) & = \frac {4x^2 + 8x + 13}{6(1 + x)}\\\\ f'(x) & = \frac{(4x^2 + 8x + 13)'(1 + x) - (4x^2 + 8x + 13)(1 + x)'}{(1 + x)^2} & = \frac{(8x + 8)(1 + x) - (4x^2 + 8x + 13)(1)}{(1 + x)^2} & = \frac{4x^2 + 8x - 5}{(1 + x)^2} f''(x) & = \frac{(4x^2 + 8x - 5)'(1 + x)^2 - (4x^2 + 8x - 5)((1 + x)^2)'}{(1 + x)^4} \end{align*}