Rational approximation of an integer combination of two irrationals Let $x$ be an irrational number, and $\beta$ strictly larger than its irrationality index, which means that for some $C>0$, for all $n\in \mathbb{Z}^*$, 
$$d(nx,\mathbb{Z})>C n^{-\beta}.$$
It is known that for a.e. irrational number $x$, the irrationality index is $1$.
It is even known that some numbers satisf the above  for $\beta=1$ (for instance, $x=\sqrt{2}$).
By arguments from measure theory, I have been able to prove that if $a_n$ satisfies $$\sum_{n=1}^{\infty}na_n<\infty,$$ almost every couple $(x,y)$ of $\mathbb{R}^2$ satisfies for some $C>0$ $$d(nx+my,\mathbb{Z})>Ca_{|n|+|m|},n,m\in \mathbb{Z}.$$
Ideally, I would like, as for a single number $x$, find irrational numbers $(x,y)$ such that this holds in the limiting case $a_n=n^{-2}$.
Has anyone an idea? Or has anyone a useful reference for such things?
 A: Yes, such $(x,y)$ exist; for example,
$x = \root 3 \of 2$ and $y = x^2$.
For $l,m,n \in \bf Z$, define
$$
N(l,m,n) := l^3 + 2m^3 + 4n^3 - 6lmn \in {\bf Z};
$$
this is the algebraic norm 
$$ 
 (l + mx + nx^2)
 (l + m\rho x + n \bar\rho x^2)
 (l + m\bar\rho x + n \rho x^2)
$$
of $l + mx + nx^2$,
where $\rho$ is the cube root of unity $e^{2\pi i/3} = (-1 + \sqrt{-3})/2$.
But if $(m,n) \neq (0,0)$ then
$l + mx + nx^2 \neq 0$, so $\left|N(l,m,n)\right| \geq 1$ and
$\left|l + mx + nx^2\right| \gg (|l|+|m|+|n|)^{-2}$.
Taking for $l$ the integer nearest to $-(mx+nx^2)$ we deduce that
$d(mx+nx^2,{\bf Z}) \gg (|m|+|n|)^{-2}$, as claimed.
The same argument (which generalizes the familiar one for $\sqrt 2$)
shows that in general if $x$ is an algebraic number of degree $D$ then
$$
d\Bigl(\sum_{j=1}^{D-1} n_j x^j, {\bf Z}\Bigr) \gg
 \left( \sum_{j=1}^{D-1} \left| n_j \right| \right)^{1-D}
$$
for $n_1,\ldots,n_{D-1} \in \bf Z$ not all zero.
This is best possible up to the value of the implicit constant, because
Dirichlet's celebrated "pigeonhole" argument shows that conversely
for any $N$ one can find integers $n_1,\ldots,n_{D-1} \in [-N,N]$, not all zero,
such that $d(\sum_{j=1}^{D-1} n_j x^j, {\bf Z}) \ll N^{1-D}$.
