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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 \(x + y\)?

 Jun 22, 2021
 #1
avatar+26112 
+2

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 \(x+y\)?

 

\(\begin{array}{|rcll|} \hline \dfrac{1}{x} + \dfrac{1}{y} &=& \dfrac{1}{18} \\\\ \dfrac{x+y}{xy} &=& \dfrac{1}{18} \\\\ \mathbf{xy} &=& \mathbf{18*(x+y)} \\ \hline \end{array}\)

 

 

\(\large{AM\ge GM}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{x+y}{2}} &\ge& \mathbf{\sqrt{xy}} \\ x+y &\ge& 2\sqrt{xy} \quad &| \quad \text{square both sides} \\ (x+y)^2 &\ge& 4xy \quad | \quad xy = 18*(x+y) \\ (x+y)^2 &\ge& 4*18*(x+y) \\ x+y &\ge& 4*18 \\ \mathbf{x+y } &\ge& \mathbf{72} \\ \hline \end{array}\)

 

\(\text{The smallest possible value for $x+y$ is $72$} \)

 

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

 

laugh

 Jun 22, 2021
 #2
avatar+26112 
+2

In general:

\(\begin{array}{|rcll|} \hline \dfrac{1}{x} + \dfrac{1}{y} &=& \dfrac{1}{n} \\\\ \mathbf{x+y } &\ge& \mathbf{4n} \\ \hline \end{array}\)

\(\text{The smallest possible value for $x+y$ is $4n$}\)

 

laugh

heureka  Jun 22, 2021

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