Simultaneous rational approximation of two reals using their continued fractions

Littlewood's well-known conjecture about simultaneous rational approximation is that for all $x, y in mathbb{R}$, $liminf_{n to infty} n Vert nx Vert Vert ny Vert = 0$ (where $Vert x Vert$ denotes the distance from $x$ to the nearest integer).

A heuristic argument for this (mentioned in this survey article by Akshay Venkatesh) is as follows. Write $q_n(x)$ for the denominator of the $n$th convergent of the continued-fraction expansion of $x$. We know that $q_n(x) Vert q_n(x) x Vert$ is bounded, and it is reasonable to expect that $Vert q_n(x) y Vert$ will be small sometimes. Venkatesh points out that this argument doesn't work: in fact, given any sequence $q_n$ such that $liminf q_{n+1} / q_n > 1$, we can find a $y in mathbb{R}$ such that $Vert q_n y Vert$ is bounded away from zero.

However, this doesn't quite rule out using the continued-fraction expansions of $x$ and $y$ to prove Littlewood's conjecture. The result would follow if it could be shown that for all $x, y in mathbb{R}$, either $liminf_{n to infty} Vert q_n(x) y Vert = 0$ or $liminf_{n to infty} Vert q_n(y) x Vert = 0$.

My question is whether this can be shown to be false. That is: do there exist $x$ and $y$ such that both $Vert q_n(x) y Vert$ and $Vert q_n(y) x Vert$ are bounded away from zero?

In light of gowers's answer below, let me add the condition that $x$ and $y$ are badly-approximable, i.e. that they have bounded partial quotients in their continued fractions. (It is still the case that a negative answer would imply Littlewood's conjecture.)

Here is an example from the realm of badly-approximable numbers. Let $x = frac12 (1 + sqrt{5})$ and $y = frac12 (1 - 1/sqrt{5})$. Then $q_n(x) = F_n$ (Fibonacci number), and $q_n(y) = 2F_n + F_{n+1}$ for $n geq 1$. In this case, $liminf Vert q_n(x) y Vert = frac15 > 0$, but $Vert q_n(y) x Vert to 0$.