1. **Problem statement:** We have a tetrahedron $ABCO$ with right angles at $O$ such that $\angle AOB = \angle AOC = \angle BOC = 90^\circ$. A cube is inscribed inside this tetrahedron with one vertex at $O$ and the opposite vertex on the face $ABC$. We want to show that the side length $s$ of the cube is $$s = \frac{abc}{ab + ac + bc}$$ where $a = OA$, $b = OB$, and $c = OC$. 2. **Setup and notation:** Since $OA$, $OB$, and $OC$ are mutually perpendicular, we can place $O$ at the origin and $A$, $B$, $C$ on the coordinate axes: $$A = (a,0,0),\quad B = (0,b,0),\quad C = (0,0,c).$$ 3. **Cube vertices and side length:** Let the cube have side length $s$ and one vertex at $O=(0,0,0)$. Because the cube edges are parallel to the coordinate axes, the opposite vertex of the cube is at $(s,s,s)$. 4. **Condition for the opposite vertex on face $ABC$:** The face $ABC$ lies in the plane passing through $A$, $B$, and $C$. The equation of plane $ABC$ can be found using the intercept form: $$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1.$$ 5. **Substitute the cube's opposite vertex into the plane equation:** Since the opposite vertex $(s,s,s)$ lies on $ABC$, it satisfies $$\frac{s}{a} + \frac{s}{b} + \frac{s}{c} = 1.$$ 6. **Solve for $s$:** Factor out $s$: $$s \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right) = 1.$$ 7. **Simplify the sum of reciprocals:** $$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{bc + ac + ab}{abc}.$$ 8. **Final expression for $s$:** $$s = \frac{1}{\frac{bc + ac + ab}{abc}} = \frac{abc}{ab + ac + bc}.$$ 9. **Interpretation:** The side length of the cube is the harmonic mean of the edges $a$, $b$, and $c$ weighted by their products, which fits the geometric constraints of the cube inscribed in the tetrahedron. This completes the proof.