I took this result from Minkowski's book on [Geometry of numbers](https://archive.org/details/geometriederzahl00minkrich):



> 
> Two arbitrary real quantitites $a$ and $b$ may be made to approach as near as we wish in value the two fractions $\frac{x}{z}$ and $\frac{y}{z}$, that have the same denominator and at the same time in such a way that:
>  $$ \left| \frac{x}{z} - a \right| < \sqrt{\frac{8}{19}} \frac{1}{z^{3/2}}
> \hspace{0.5in}\left| \frac{y}{z} - b \right| < \sqrt{\frac{8}{19}} \frac{1}{z^{3/2}}$$
> 
> 
> 


I interpret this as we are trying to approximate *both* $a, b \in \mathbb{R}$ with two fractions $\frac{x}{z},\frac{y}{z} \in \mathbb{Q}$ with the same denominator to $O(z^{-3/2})$.


How to prove this result? Is $\sqrt{\frac{8}{19}}$ optimal? 


I will settle for worse constant just to see how geometry of numbers works here.



approximate two different real numbers to order $\frac{1}{z^{3/2}}$
With the constant $1$, this is Minkowski's higher dimensional extension of Dirichlet's approximation theorem:


*If $\alpha\_1, \ldots,\alpha\_n$ are real numbers, then there are rationals $p\_i/q$ with $|\alpha\_i - p\_i/q| < q^{-(1+1/n)}$. If one at least among the $\alpha\_i$ is irrational, there are infinitely many such $n$-tuples $(p\_1/q,\ldots,p\_n/q)$.* 


The proof is a simple pigeonholing. With an integer parameter $Q$, consider the $Q^n+1$ points $(1,\ldots,1)$ and $(\{\alpha\_1 x\}, \ldots,\{\alpha\_n x\})$, $0 \leq x < Q^n$ integer, of the unit cube $[0,1]^n$. Some two such points must be contained by a common cube of side $1/Q$. Taking their difference produces a positive integer $q < Q^n$ and integers $p\_1, \ldots, p\_n \in \mathbb{Z}$ (arising as the appropriate integer parts) with $|q\alpha\_i - p\_i| \leq 1/Q$ for all $i = 1,\ldots,n$. We have $1/Q < q^{-1/n}$, giving the inequality. Moreover, we may by clearing the common factor assume $(p,q\_1,\ldots,q\_n) = 1$ for the tuple thus produced, and if say $\alpha\_1$ is irrational it is clear that for a fixed tuple $|q\alpha\_1 - p\_1| \leq 1/Q$ can hold only for finitely many $Q$. Hence, letting $Q \to \infty$, infinitely many tuples $(q;p\_1,\ldots,p\_n)$ are thus obtained when some $\alpha\_i \notin \mathbb{Q}$.




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Better constants can be obtained. For $n = 1$ the optimal constant is $1/\sqrt{5}$ as you know (and then you have the Lagrange-Markov spectrum). For $n = 2$ the optimal constant is unknown as far as I am concerned; this is stated on page 41 of Wolfgang Schmidt's book "Diophantine Approximation" (LNM 785), to which I can refer you for an improvement of the constant $1$ in the general case. There, in particular, the value $2/3$ is shown to work for $n = 2$ (which however is slightly bigger than the constant you cite from Minkowski book), and a reference is given to the literature proving that the optimal constant lies between the values $\sqrt{2/7} \approx 0.53$ and $0.615 < 0.649 \approx \sqrt{8/19}$. 


So no, the constant $\sqrt{8/19}$ is not optimal in this result.


