= Solution
Let $P_\phi=|\phi\rangle\langle\phi|$. An arbitrary deterministic strategy, including auxiliary systems, intermediate measurements, adaptive actions and discarded records, is a <quantum channel> $\mathcal E$ from the two input qubits to the three returned qubits. Its output is $R_\phi=\mathcal E(P_\phi^{\otimes2})$. The three local tests commute, and their joint all-pass projector is $P_\phi^{\otimes3}$. Thus
$$
\epsilon_\phi=1-\operatorname{Tr}(P_\phi^{\otimes3}R_\phi)
$$
is exactly the <probability> of at least one failed test. No independence of the three returned qubits is assumed.
We prove a uniform positive obstruction using <trace distance>, $D(\rho,\sigma)=\|\rho-\sigma\|_1/2$. Two normalized pure kets obey
$$
D(|u\rangle\langle u|,|v\rangle\langle v|)=\sqrt{1-|\langle u|v\rangle|^2}.
$$
Indeed their projector difference is supported on their two-dimensional span, has trace zero, and has <eigenvalues> $\pm\sqrt{1-|\langle u|v\rangle|^2}$.
Two further estimates will cover even mixed outputs. First, <trace distance> contracts under a <quantum channel>. To see this directly, the positive spectral projector of the trace-zero difference gives $D(\rho,\sigma)=\max_{0\leq A\leq I}\operatorname{Tr}[A(\rho-\sigma)]$. The adjoint of a channel is positive and unital, so it takes any output effect into an input effect. Maximizing over this subset cannot exceed the input maximum, proving contraction.
Second, for a mixed state $R=\sum_jq_j|u_j\rangle\langle u_j|$ and a pure target $P=|v\rangle\langle v|$, the trace-norm triangle inequality and weighted <Cauchy-Schwarz inequality> give the <pure-target upper bound on trace distance>:
$$
D(R,P)\leq\sum_jq_j\sqrt{1-|\langle u_j|v\rangle|^2}
\leq\sqrt{1-\langle v|R|v\rangle}.
$$
In particular $D(R_\phi,P_\phi^{\otimes3})\leq\sqrt{\epsilon_\phi}$.
Now take $|\phi_0\rangle=|0\rangle$ and $|\phi_1\rangle=|+\rangle=(|0\rangle+|1\rangle)/\sqrt2$. The two-copy input distance is $D_2=\sqrt{1-1/4}=\sqrt3/2$, whereas the ideal three-copy distance is $D_3=\sqrt{1-1/8}=\sqrt{7/8}$. Writing $R_j=R_{\phi_j}$, contraction and the triangle inequality yield
$$
D_3\leq D(P_{\phi_0}^{\otimes3},R_0)+D(R_0,R_1)+D(R_1,P_{\phi_1}^{\otimes3})
\leq\sqrt{\epsilon_0}+D_2+\sqrt{\epsilon_1}.
$$
Let $\delta=D_3-D_2>0$. Then $\sqrt{\epsilon_0}+\sqrt{\epsilon_1}\geq\delta$, and squaring gives
$$
\boxed{\frac{\epsilon_0+\epsilon_1}{2}\geq\frac{\delta^2}{4}>0.}
$$
This is the <copy-number trace-distance obstruction>. It applies to every Bob strategy, irrespective of its internal complexity. If Alice chooses these two possible inputs with equal <probabilities>, the left side is her unconditional failure <probability>. For every device, at least one of the two inputs also has failure <probability> at least $\delta^2/4$, establishing the worst-case guarantee.
The same positive bound holds if Alice chooses a uniformly random pure-qubit state. Apply the preceding inequality to $U|0\rangle$ and $U|+\rangle$ for every single-qubit unitary $U$, since their overlap is unchanged. Average over normalized <Haar measure>. Both rotated kets have the same uniform distribution on the <Bloch sphere>, so each of the two average failures is the uniform-input average. Consequently that average is also at least $\delta^2/4$. A constant satisfying the strict requested inequality in either this average or the worst-case interpretation is
$$
\boxed{p=\frac18\left(\sqrt{\frac78}-\frac{\sqrt3}{2}\right)^2>0.}
$$
The proven bound is $2p$, hence strictly greater than $p$. This proof establishes a quantitative gap, rather than relying only on the impossibility of exact <quantum cloning>.
There is a necessary quantifier qualification because the PDF does not specify Alice's input distribution. A positive lower bound for every fixed state against every strategy would be false: a device hardwired to return $|000\rangle$ never fails when the input is $|0\rangle^{\otimes2}$. The unknown-state assertion is therefore meaningful as a worst-case guarantee over possible inputs, or as an average for a specified hidden input ensemble. Both the equal-prior example and the uniform-input version have been proved here, with a constant independent of Bob's strategy.
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