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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
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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 $$x+y$$?

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

$$\large{AM\ge GM}$$

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

$$\text{The smallest possible value for x+y is 72}$$ Jun 22, 2021
#2
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In general:

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

$$\text{The smallest possible value for x+y is 4n}$$ heureka  Jun 22, 2021