Given a real number $\alpha$ we can define its irrationality measure $\mu=\mu(\alpha)$ as the largest number such that
$$\inf_{|a|, |b|\leq n} |a+b \alpha| \leq \frac{k}{n^{\mu+\epsilon}}$$
for some constant $k$ for all $n\in\mathbb N$ and $\epsilon>0$. This definition is shifted by $1$ from the traditional one, but that won't matter for my purposes. We can also define the irrationality measure of a pair of real numbers $\alpha, \beta$ as the largest $\mu=\mu(\sqrt 2, \sqrt 3)$ such that
$$\inf_{|a|, |b|, |c|\leq n} |a+b \alpha+c\beta| \leq \frac{k}{n^{\mu+\epsilon}}.$$
I want to know what is the irrationality measure of the most obvious pair of numbers to try, $\sqrt 2$ and $\sqrt 3$. We have that
\begin{align} \inf |a+b\sqrt 2+c\sqrt 3|&=\inf \frac{|a^4 - 4 a^2 b^2 - 6 a^2 c^2 + 4 b^4 - 12 b^2 c^2 + 9 c^4|}{|a-b\sqrt 2+c\sqrt 3||a+b\sqrt 2-c\sqrt 3||a-b\sqrt 2-c\sqrt 3|}\\ &\geq \inf \frac 1{|a-b\sqrt 2+c\sqrt 3||a+b\sqrt 2-c\sqrt 3||a-b\sqrt 2-c\sqrt 3|}\\ &\geq \frac k{n^3}. \end{align} The manipulations are justified because the square roots of the primes are linearly independent over the rationals and all the three terms in the denominator have to be comparable to $n$ or you just get an approximation to $\sqrt 2$ or $\sqrt 3$ which can only be of the order $\frac 1{n^2}$. So, what is $\mu(\sqrt 2, \sqrt 3)$?