# 1965 AHSME Problems/Problem 34

## Contents

## Problem 34

For the smallest value of is:

## Solution 1

To begin, lets denote the equation, as . Let's notice that:

After this simplification, we may notice that we may use calculus, or the AM-GM inequality to finish this problem because , which implies that both are greater than zero. Continuing with AM-GM:

Therefore, ,

## Solution 2 (Calculus)

Take the derivative of f(x) and f'(x) using the quotient rule.

## Solution 3 (answer choices, no AM-GM or calculus)

We go from A through E and we look to find the smallest value so that , so we start from A:

However by the quadratic formula there are no real solutions of , so cannot be greater than 0. We move on to B:

There is one solution: , which is greater than 0, so 2 works as a value. Since all the other options are bigger than 2 or invalid, the answer must be