Let D be the vertex of the cube that lies on face ABC and E, F, G be the vertices of the cube adjacent to D.
Since AOB is a right angle, we have AD^2 + BD^2 = AB^2. Similarly, we have AD^2 + CD^2 = AC^2 and BD^2 + CD^2 = BC^2.
Also, since D is a vertex of the cube, we have DE = DF = DG. Let x be the length of the side of the cube. Then we have AD = AE = AF = a - x, BD = BE = BG = b - x, and CD = CF = CG = c - x.
Using the Pythagorean theorem in triangles ADE, BDE, and CDE, we have
(x - a)^2 + DE^2 = AD^2 (x - b)^2 + DE^2 = BD^2 (x - c)^2 + DE^2 = CD^2
Substituting DE = DF = DG = x/√2 and simplifying, we get
(x - a)^2 + (x^2/2) = (a - x)^2 (x - b)^2 + (x^2/2) = (b - x)^2 (x - c)^2 + (x^2/2) = (c - x)^2
Expanding the squares and simplifying, we get
3x^2 = a^2 + b^2 + c^2 - 2ab - 2ac - 2bc
Solving for x, we get
x = √(2abc(a + b + c))/(2(ab + ac + bc))
Substituting this value of x in the expressions for AD, BD, and CD, we get the side lengths of the cube:
AD = a - √(2abc(a + b + c))/(2(ab + ac + bc)) BD = b - √(2abc(a + b + c))/(2(ab + ac + bc)) CD = c - √(2abc(a + b + c))/(2(ab + ac + bc))
The length of the side of the cube is the minimum of these three lengths, so we have
x = min{AD, BD, CD} = a - √(2abc(a + b + c))/(2(ab + ac + bc))
Therefore, the side length of the cube is
abc/(ab + ac + bc).
Q.E.D.
