I've been trying to prove this
Given any $x,y \in \mathbb{N}$ with $x < y$, $\dfrac{2xy}{x+y} \not\in \mathbb{N}$.
Here's what I tried:
For contradiction, suppose there were some $x,y \in \mathbb{N}$ such that $\dfrac{2xy}{x+y}\in \mathbb{N}$. Let $\dfrac{2xy}{x+y}=H$. Then $y\left( 2\dfrac{x}{H}-1 \right)= x$. Since $H$ is the harmonic mean of $x$ and $y$, we have $0<x<H<y$. Now, $0< \dfrac{x}{H} < 1$, thus, $\dfrac{x}{H} \not\in \mathbb{N}$. Hence, we have $x= y\left( 2\dfrac{x}{H}-1 \right) \not \in \mathbb{N}$. This leads to contradicts the fact that $x$ was in $\mathbb{N}$. This ends the proof.
Is my proof correct? What would be other way to do it?
Your proof can't be right, since what you're trying to prove is not true.
In particular $(x,y)=(3,6)$ is a counterexample: $\frac{2\cdot 3\cdot 6}{3+6}=\frac{36}{9}=4\in\mathbb N$.
For this counteexample it is true that $\frac xH = \frac 34$ is not a natural number. And $2\frac xH-1 = \frac12$ is not a natural number either. But when we multiply that with $6$, the denominator disappears.