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 $(\mathbf{t,u,v})$ for the sides satisfying: $$\mathbf{t.t}=a^2, \mathbf{u.u}=b^2, \mathbf{v.v}=c^2,$$ $$\mathbf{t.u}=0,\phantom{^2} \mathbf{u.v}=0,\phantom{^2} \mathbf{v.t}=0,\phantom{^2}$$ $$\mathbf{\pm t\pm u\pm v} \le (x,y,z)$$ 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.