Solution
= Solution
Work in the rest frame of $B$. Two-body decay kinematics gives the massless-particle energies
$$
E_1=\frac{(m_A^2-m_B^2)c^2}{2m_B},
\qquad
E_2=\frac{(m_B^2-m_C^2)c^2}{2m_B}.
$$
If $\theta$ is the angle between their momenta, then
$$
(Q_1+Q_2)^2=2Q_1\cdot Q_2
=\frac{2E_1E_2}{c^2}(1-\cos\theta).
$$
Since $0\leq1-\cos\theta\leq2$,
$$
\boxed{0\leq(Q_1+Q_2)^2\leq
\frac{(m_A^2-m_B^2)(m_B^2-m_C^2)c^2}{m_B^2}.}
$$