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Consider the tetrahedron with face areas being proportional to $1, 1, \frac{1}{2}, \frac{1}{3}$ (such a tetrahedron exists because the sum of any three of these numbers is greater than the fourth one). Then its altitudes will be proportional to $1, 1, 2, 3$. Obviously no three of them are sides of a triangle.

On the competition, you had better hope the answer was "no" and find a counter example because proving "yes" would be quite a task (impossible in this case).