4 fixed spelling of Akshay's name

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 Arkshay 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 $\Vert q_n(y) x \Vert \to 0$.

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 Arkshay 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?

EDIT:

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 $\Vert q_n(y) x \Vert \to 0$.

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 Arkshay 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?
EDIT: 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.)