Given positive integers \(x\) and \(y\) such that \(x\neq y\) and \(\frac{1}{x} + \frac{1}{y} = \frac{1}{18} \), what is the smallest possible value for \(x + y\)?
+26367 Given positive integers \(x\) and \(y\) such that \(x \ne y\) and \(\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{18}\),
what is the smallest possible value for \(x+y\)?
\(\begin{array}{|rcll|} \hline \dfrac{1}{x} + \dfrac{1}{y} &=& \dfrac{1}{18} \\\\ \dfrac{x+y}{xy} &=& \dfrac{1}{18} \\\\ \mathbf{xy} &=& \mathbf{18*(x+y)} \\ \hline \end{array}\)
\(\large{AM\ge GM}\)
\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{x+y}{2}} &\ge& \mathbf{\sqrt{xy}} \\ x+y &\ge& 2\sqrt{xy} \quad &| \quad \text{square both sides} \\ (x+y)^2 &\ge& 4xy \quad | \quad xy = 18*(x+y) \\ (x+y)^2 &\ge& 4*18*(x+y) \\ x+y &\ge& 4*18 \\ \mathbf{x+y } &\ge& \mathbf{72} \\ \hline \end{array}\)
\(\text{The smallest possible value for $x+y$ is $72$} \)
