= Solution
Addition of two spin-one <angular momenta> gives $J=0,1,2$, with $M=-J,-J+1,\ldots,J$. To obtain the $J=1$ multiplet, its highest-weight state must have $M=1$ and be killed by $J_+=J_+^{(1)}+J_+^{(2)}$. On the two possible product states, the raising operator gives the same multiple of $|1;1\rangle$, so their difference is required. Normalization and successive lowering give
$$
\boxed{\begin{aligned}
|1,1\rangle&=\frac{|1;0\rangle-|0;1\rangle}{\sqrt2},\\
|1,0\rangle&=\frac{|1;-1\rangle-|-1;1\rangle}{\sqrt2},\\
|1,-1\rangle&=\frac{|0;-1\rangle-|-1;0\rangle}{\sqrt2}.
\end{aligned}}
$$
For example $J_-|1,1\rangle=|1;-1\rangle-|-1;1\rangle=\sqrt2|1,0\rangle$, and lowering once more gives the last expression. These states are all antisymmetric under interchanging the two constituents.
Identical spin-one particles are <bosons>, so the total state must be symmetric. Its $J=1$ spin factor is antisymmetric and must therefore multiply an antisymmetric spatial factor. Two particles in the same one-particle orbital have zero antisymmetric spatial state. The lowest allowed choice is
$$
\frac{\psi_1(r_A)\psi_2(r_B)-\psi_2(r_A)\psi_1(r_B)}{\sqrt2}\,|1,M\rangle,
\qquad \boxed{E=E_1+E_2},\quad M=-1,0,1.
$$
For distinguishable particles there is no exchange-symmetry constraint. Both can occupy the ground orbital, giving $\psi_1(r_A)\psi_1(r_B)|1,M\rangle$ and \b[lowest energy $\boxed{2E_1}$], again with three spin states.
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