Let $f(x)=\frac{A}{x+B}$ for $A,B\in \mathbb{R}$. Let $\alpha_1,\alpha_2$ be linearly independent over $\mathbb{Q}$. Is it true that $$|f(\mathbb{Q})\cap (\alpha_1\cdot[1,t]+\alpha_2\cdot [1,t])|\le Kt$$ for some universal constant $K$ (not depending on $A$ or $B$)?
I know this is true for $f(x)=a_nx^n+\dots+a_1x^1+a_0$. By the Pigeonhole Principle, if we have $|f(\mathbb{Q})\cap (\alpha_1\cdot[1,t]+\alpha_2\cdot [1,t])|> nt$, then $f(\mathbb{Q})\ni \alpha_1\cdot y+\alpha_2\cdot z$ for some fixed value $y$ and at least $n+1$ distinct values of $z$. It follows that $|f(\mathbb{Q})\cap (\alpha_1 y+\alpha_2\cdot \mathbb{Q})|\ge n+1$.
Let $g(x)=f(x)-\alpha_1\cdot y$. Then $|g(\mathbb{Q})\cap\alpha_2\cdot \mathbb{Q}|\ge n+1$, so there are distinct $x_1,\dots,x_{n+1}\in \mathbb{Q}$ so that $g(x_i)\in \alpha_2\cdot \mathbb{Q}$ for all $1\le i\le n+1$. Letting $V$ be the Vandermonde matrix $$V=\begin{pmatrix} 1&x_1&\cdots &x_1^{n}\\ 1&x_2&\cdots &x_2^{n}\\ \vdots&\vdots&\ddots&\vdots\\ 1&x_{n+1}&\cdots&x_{n+1}^n \end{pmatrix}$$ We have $$V\begin{pmatrix}a_0-\alpha_1y\\ a_1 \\ \vdots \\ a_n \end{pmatrix}\in\alpha_2\cdot \mathbb{Q}^{n+1}$$ Since $V$ is a matrix in $\mathbb{Q}^{n+1\times n+1}$ and is invertible, it follows that $$\begin{pmatrix}a_0-\alpha_1y\\ a_1 \\ \vdots \\ a_n \end{pmatrix}\in\alpha_2\cdot V^{-1}\mathbb{Q}^{n+1}\subseteq\alpha_2\cdot\mathbb{Q}^{n+1}$$ Hence every coefficient of $g$ is in $\alpha_2\cdot\mathbb{Q}$, so $g(\mathbb{Q})\subseteq \alpha_2\cdot\mathbb{Q}$. So $g(\mathbb{Q})\subseteq \alpha_1\cdot y+\alpha_2\cdot\mathbb{Q}$, so $|f(\mathbb{Q})\cap (\alpha_1\cdot[1,t]+\alpha_2\cdot [1,t])|\le 2n$.
I've tried adapting this idea to the case $f(x)=\frac{A}{x+B}$ for real $A,B$, but I've had no luck so far. Any ideas?