= Solution
Put $J=j_1+j_2$. The only state with $m=J$ is the maximal <highest-weight states in angular momentum addition>[highest-weight state]
$$
|J,J\rangle=|j_1,j_1\rangle_1|j_2,j_2\rangle_2.
$$
Applying $J_-=J_{1-}+J_{2-}$ and normalizing gives
$$
|J,J-1\rangle
=\sqrt{\frac{j_1}{J}},
|j_1,j_1-1\rangle_1|j_2,j_2\rangle_2
+\sqrt{\frac{j_2}{J}},
|j_1,j_1\rangle_1|j_2,j_2-1\rangle_2.
$$
The orthonormal vector in the same two-dimensional $m=J-1$ subspace is
$$
|J-1,J-1\rangle
=\sqrt{\frac{j_2}{J}},
|j_1,j_1-1\rangle_1|j_2,j_2\rangle_2
-\sqrt{\frac{j_1}{J}},
|j_1,j_1\rangle_1|j_2,j_2-1\rangle_2.
$$
It is annihilated by $J_+$, so it is the highest-weight state of the $j=J-1$ multiplet. These are all states with $m\geq J-1$. The last state exists when both subsystem spins are nonzero.
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