= Solution
The same <Super-Poincaré algebra> gives the operator identity
$$
H=\frac14\sum_{\alpha=1}^2\{Q_\alpha,Q_\alpha^\dagger\}.
$$
For a normalized physical state $|v\rangle$ in a positive-norm <Hilbert space>,
$$
\boxed{\langle H\rangle_v=\frac14\sum_{\alpha=1}^2\big(\|Q_\alpha v\|^2+\|Q_\alpha^\dagger v\|^2\big)\ge0.}
$$
An unbroken supersymmetric vacuum must be annihilated by every <supercharge> and its adjoint, hence has zero energy. Conversely, zero expectation of this positive sum forces each <norm> separately to vanish. Therefore a vacuum with \b[strictly positive energy necessarily breaks <supersymmetry>]. For an infinite-volume field theory the corresponding statement is about <vacuum energy density>, obtained with an appropriate finite-volume regulator. The zero of energy is fixed by the algebra; it cannot be arbitrarily shifted while preserving this identity. This is the global-supersymmetry result and does not assert positivity of the general local-supergravity scalar potential.
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