10.3. First, let's assume that if $\angle B A C=60^{\circ}$, then $A B + A C \leqslant 2 B C$. For this, consider points $B^{\prime}$ and $C^{\prime}$, which are symmetric to points $B$ and $C$ relative to the bisector of angle $A$. Since in any convex quadrilateral, the sum of the lengths of the diagonals is greater than the sum of the lengths of a pair of opposite sides, we have $B C + B^{\prime} C^{\prime} \geqslant C C^{\prime} + B B^{\prime}$ (equality is achieved if $A B = A C$). It remains to note that $B^{\prime} C^{\prime} = B C$, $C C^{\prime} = A C$, and $B B^{\prime} = A B$. Similarly, the inequalities $A C + A D \leqslant 2 C D$ and $A D + A B \leqslant 2 D B$ can be proven. Adding all these inequalities, we obtain the required result.