# Difference between revisions of "1982 AHSME Problems/Problem 29"

## Problem

Let $x,y$, and $z$ be three positive real numbers whose sum is $1$. If no one of these numbers is more than twice any other, then the minimum possible value of the product $xyz$ is

$\textbf{(A)}\ \frac{1}{32}\qquad \textbf{(B)}\ \frac{1}{36}\qquad \textbf{(C)}\ \frac{4}{125}\qquad \textbf{(D)}\ \frac{1}{127}\qquad \textbf{(E)}\ \text{none of these}$

## Solution

The answer is A, 1/32, as obtained by (1/4) * (1/4) * (1/2).