= Cloning failure bound from trace-distance contraction
{title2=$\sqrt{e_0}+\sqrt{e_1}\geq\sqrt{1-c^{2N}}-\sqrt{1-c^{2M}}$}
For two <pure states> of overlap $0<c<1$, a <quantum channel> attempts to turn $M$ copies into $N>M$ copies. Let $e_i$ be the failure probability of the all-copies projector test on input $i$. The <pure-state trace distance and fidelity identity>, <trace-distance contraction under quantum channels>, <pure-target upper bound on trace distance> and <triangle inequality> give the displayed strictly positive gap $\Delta$. Hence $\max_i e_i\geq\Delta^2/4$, and for positive priors $\pi_0+\pi_1=1$, weighted <Cauchy-Schwarz inequality> gives $\pi_0e_0+\pi_1e_1\geq\pi_0\pi_1\Delta^2$. No positive state-by-state bound holds: preparing one candidate perfectly makes its own conditional failure probability zero.
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