For each positive real number $x$ let $S(x)=\{\lfloor kx\rfloor \,:\, k \in\mathbb{N}\}$. Let $x_{1},x_{2},x_{3}$ be positive real numbers each greater than $1$ such that $\sum_{i=1}^{3}\frac{1}{x_i} >1$. Prove that there exists $i$ and $j$ (where $1\leq i<j\leq 3$) such that $|S(x_i)\cap S(x_j)| =\infty$.
This problem looks hard and i have no idea how to start.
Write $[N] = \{1, \ldots, N\}$. Then note that $$\#(S(x) \cap [N]) \geq N/x + \mathcal O(1)$$ for fixed $x > 1$. Write $\alpha = 1/x_1 + 1/x_2 + 1/x_3 > 1$. It follows that $$ \#(S(x_1) \cap [N]) + \#(S(x_2) \cap [N]) + \#(S(x_3) \cap [N]) \geq \alpha N + \mathcal O(1). $$ As $N$ tends to infinity the gap between the sum on the left-hand side and $N$ grows without bound. Thus at least one pair-wise intersection between the $S(x_i)$ has to be infinite.