This could also be solved by quantifier elimination, and perhaps someone with more knowledge of quantifier elimination can see how to do it feasibly.
A box of side lengths $(a,b,c)$ fits inside a box of side lengths $(x,y,z)$ iff there are vectors $(\mathbf{t},\mathbf u,\mathbf{v})$ for the sides satisfying: \begin{gather*} \begin{aligned} \mathbf{t}\cdot\mathbf{t}=a^2,&& & \mathbf{u}\cdot\mathbf{u}=b^2,&& \mathbf{v}\cdot\mathbf{v}=c^2, \\ \mathbf{t}\cdot\mathbf{u}=0,&&& \mathbf{u}\cdot\mathbf{v}=0,&& \mathbf{v}\cdot\mathbf{t}=0, \end{aligned} \\ \pm\mathbf{t} \pm \mathbf{u} \pm \mathbf{v} \le (x,y,z) \end{gather*} where the last line represents three coordinate inequalities for each choice of signs. In other words, we are checking a statement of the form $\exists t_i t_j t_k u_i u_j u_k v_i v_j v_k\ \phi$, where $\phi$ is the conjunction of 6 equalities and 24 inequalities in those 9 bound variables and in $a$, $b$, $c$, $x$, $y$, $z$.
Quantifier elimination should then provide some list of polynomial inequalities in $a$, $b$, $c$, $x$, $y$, $z$ which together are equivalent to the small box fitting inside the large box.
Update: This works in the 2-d case, with interesting results. Assuming that $0<a<b$ and $0<x<y$, the $a\times b$ rectangle can fit inside the $x\times y$ rectangle iff either: $$a \le x$$ $$b \le y$$ or: $$a \le x$$ $$a^2+b^2 \le x^2+y^2$$ $$\phantom{(ax+by)^2+}(b^2-a^2)^2 \le (ax-by)^2+(ay-bx)^2$$