I saw this problem in the prep book for the USAMO in the Algebra section, but I don’t know how to tackle it. As the problems before were all solvable by factoring/Cauchy-Schwarz, I have tried finding a way to factor/rearrange this into a C.S.-form , but I have made no progress at all. As I’m getting quite frustrated with this problem I would appreciate all help!
Let $x$ and $y$ be positive integers and $k$ be a positive real number for which $$\frac{x + 1}{y + k} = \frac{y}{x}$$ is satisfied.
Prove that $k\geq1$ holds.
My thoughts: After rearranging we arrive at $(x+1)x$ which must at least be 2. And $y(y+k)$ where $y$ is at least $1$. So we have $$x^2+x=y^2+yk$$ and $$\frac{x^2+x-y^2}{y}=k$$ This obviously holds for $x>y$, but for $y<x$ I am running into difficulties.
\begin{align*} & \frac{a^2+a-b^2}{b}=c \\[4pt] \implies\;& \frac{a^2+a-b^2}{b} > 0 \\[4pt] \implies\;& a^2+a-b^2 > 0 \\[4pt] \implies\;& a^2+a > b^2 \\[4pt] \implies\;& (a+1)^2 > b^2 \\[4pt] \implies\;& a+1 > b \\[4pt] \implies\;& a\ge b\;\;\;\text{$\bigl($for integers $x,y$,$\;$if $x+1 >y\;$then $x\ge y\bigr)$} \\[4pt] \implies\;& c=\frac{a^2+a-b^2}{b}\ge \frac{b^2+b-b^2}{b}=1 \\[4pt] \end{align*}