= Solution
Let $\Gamma=(-1)^F$ be <fermion parity>. It acts as $+1$ on <bosons> and $-1$ on <fermions>, and anticommutes with every odd <supercharge>. In a finite physical <supermultiplet> at fixed positive energy $E$, trace cyclicity gives
$$
\operatorname{Tr}\bigl[\Gamma\{Q_\alpha,Q_\alpha^\dagger\}\bigr]=0.
$$
Indeed, cycling $Q_\alpha$ in the second term and using $Q_\alpha\Gamma=-\Gamma Q_\alpha$ cancels the first term. Summing diagonal spinor components of the <Super-Poincaré algebra> gives $\sum_\alpha\{Q_\alpha,Q_\alpha^\dagger\}=4H$, since the Pauli matrices are traceless. Hence <supertrace pairing at positive energy> yields
$$
4E\operatorname{Tr}\Gamma=4E(n_B-n_F)=0,\qquad\boxed{n_B=n_F\quad(E>0).}
$$
This is <boson-fermion degeneracy in a supermultiplet>. Equivalently, a nonzero charge combination with positive anticommutator gives an invertible parity-changing map between the two sectors. The positive-energy qualification is important: a zero-energy supersymmetric vacuum may be a one-state bosonic representation and need not have a fermionic vacuum partner.
For any normalizable state $|v\rangle$, the same algebra gives <energy positivity in global supersymmetry>:
$$
\langle v|H|v\rangle=\frac14\sum_{\alpha=1}^2\left(\|Q_\alpha|v\rangle\|^2+\|Q_\alpha^\dagger|v\rangle\|^2\right)\ge0.
$$
It vanishes exactly when every supercharge and its adjoint annihilates the state. In particular, for an existing translation-invariant vacuum, this proves
$$
\boxed{E_{\rm vac}=0\iff Q_\alpha|0\rangle=\bar Q_{\dot\alpha}|0\rangle=0\ \text{for all components};\qquad\text{spontaneously broken global SUSY}\iff E_{\rm vac}>0.}
$$
In infinite volume, interpret the statement as one about vacuum energy density, using finite-volume normalization first. The energy zero is fixed by the supersymmetry algebra; adding an arbitrary constant to the Hamiltonian while leaving that algebra unchanged is not allowed. For a canonical matter and gauge action, the same positivity is seen from $V=\sum_i|F_i|^2+\tfrac12\sum_aD_a^2$, so a nonzero auxiliary-field expectation breaks the symmetry. This global Minkowski statement does not apply unchanged to the <supergravity F-term potential>, which can have negative-energy supersymmetric vacua.
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