+26388 Given positive integers \(x\) and \(y\) such that \(x\neq y\) and \(\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{20}\),
what is the smallest possible value for \(x + y\)?
\(\begin{array}{|rcll|} \hline \dfrac{1}{x} + \dfrac{1}{y} &=& \dfrac{1}{20} \\\\ \dfrac{x+y}{xy} &=& \dfrac{1}{20} \\\\ \mathbf{xy} &=& \mathbf{20*(x+y)} \\ \hline \end{array}\)
\(\large{AM\ge GM} \)
\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{x+y}{2}} &\ge& \mathbf{\sqrt{xy}} \\ x+y &\ge& 2\sqrt{xy} \quad &| \quad \text{square both sides} \\ (x+y)^2 &\ge& 4xy \quad | \quad xy = 20*(x+y) \\ (x+y)^2 &\ge& 4*20*(x+y) \\ x+y &\ge& 4*20 \\ \mathbf{x+y } &\ge& \mathbf{80} \\ \hline \end{array}\)
\(\text{The smallest possible value for $x+y$ is $80$}\)
Source: https://www.quora.com/Given-positive-integers-x-and-y-x-does-not-equal-y-and-frac-1-x-frac-1-y-frac-1-12-what-is-the-smallest-possible-value-for-x-y
In general:
\(\begin{array}{|rcll|} \hline \dfrac{1}{x} + \dfrac{1}{y} &=& \dfrac{1}{n} \\\\ \mathbf{x+y } &\ge& \mathbf{4n} \\ \hline \end{array}\\ \text{The smallest possible value for $x+y$ is $4n$}\)
