approximate two different real numbers to order $\frac{1}{z^{3/2}}$ I took this result from Minkowski's book on Geometry of numbers:

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.
 A: 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 $\alpha_i \notin \mathbb{Q}$.

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.
A: This is a supplement to Vesselin Dimitrov's answer. For $n=2$ the infimum of admissible constants is between $(2/7)^{1/2}\approx 0.5345$ and $8/13\approx 0.6154$, and it has been conjectured that $(2/7)^{1/2}$ is the truth. The lower bound is due to Cassels (J. London Math. Soc. 30, (1955), 119-121), the upper bound is due to Nowak (Manuscripta Math. 36 (1981/82), 33-46).
P.S. Schmidt's book "Diophantine approximation" on page 41 refers to Mack's earlier result, which was slightly improved by Nowak. Schmidt is somewhat misleading to give $0.615$ as an upper bound, because Mack's upper bound was $\approx 0.6155$, and Nowak's upper bound is $\approx 0.6154$, both bigger than $0.615$.
