= Solution
The resource here is one copy of a <two-qubit Werner state>, with the physical range $1/2<F\leq1$. The impossibility must include successful conditional branches of <local operations and classical communication>; average monotonicity alone would not exclude probabilistic enhancement.
For a normalized two-<qubit> <pure state> $|\chi\rangle=\sum_{i,j}M_{ij}|ij\rangle$, define its <concurrence> as $C(\chi)=2|\det M|$. Its <mixed-state concurrence of two qubits> is the <convex roof extension>
$$
C(\rho)=\inf_{\rho=\sum_j r_j|\chi_j\rangle\langle\chi_j|}
\sum_jr_jC(\chi_j).
$$
Combining near-optimal decompositions proves convexity. A <separable quantum state> has zero <concurrence>, because it has a decomposition into product vectors with zero coefficient determinant.
First establish the <concurrence of a two-qubit Werner state>. A <pure state> with squared <Schmidt coefficients> $\lambda,1-\lambda$ has $C=2\sqrt{\lambda(1-\lambda)}$. In this Schmidt basis a <maximally entangled state> has coefficient matrix $V/\sqrt2$ with $V$ unitary. Its overlap magnitude is at most $(\sqrt\lambda+\sqrt{1-\lambda})/\sqrt2$, because $|V_{ii}|\leq1$. Thus the singlet overlap satisfies
$$
|\langle\Psi_-|\chi\rangle|^2\leq\frac{1+C(\chi)}2.
$$
Averaging any decomposition of $W_F$ gives $C(W_F)\geq2F-1$.
For the reverse bound use the orthonormal triplet vectors
$$
t_1=\Psi_+,\qquad t_2=\Phi_-,\qquad t_3=i\Phi_+,
\qquad
|\chi_{j,\pm}\rangle=\sqrt F\,|\Psi_-\rangle
\pm\sqrt{1-F}\,|t_j\rangle.
$$
Each is normalized, and direct evaluation of its two-by-two coefficient determinant gives $C(\chi_{j,\pm})=2F-1$. Their equally weighted six-state mixture has cancelled singlet–triplet coherences and triplet weight $(1-F)/3$, so it is exactly $W_F$. Hence
$$
\boxed{C(W_F)=2F-1.}
$$
Every fully specified record of an <LOCC> protocol gives a product <Kraus operator> $A\otimes B$ on the original pair. Local ancillas and discarded local outcomes can be included by refining the record. Write its success probability and normalized output as
$$
p=\operatorname{Tr}[(A\otimes B)W_F(A^\dagger\otimes B^\dagger)],
\qquad
\rho'=\frac{(A\otimes B)W_F(A^\dagger\otimes B^\dagger)}p.
$$
For invertible $A,B$, the <concurrence scaling under invertible local filters> is
$$
C(\rho')=\frac{|\det A\det B|}{p}C(W_F).
$$
Here is a derivation of that scaling. Each pure coefficient matrix transforms to $AMB^{\mathsf T}$, so its unnormalized determinant is multiplied by $\det A\det B$. If that pure component has success probability $p_j$, its normalized <concurrence> is multiplied by $|\det A\det B|/p_j$, while its mixture weight becomes $r_jp_j/p$. Every ensemble average therefore scales by $|\det A\det B|/p$. Invertibility provides the reverse correspondence for every decomposition of $\rho'$, so taking both infima proves equality. If either local filter has rank one, the output is a <separable quantum state> and the desired bound is immediate.
It remains to show that the filter normalization cannot make this ratio exceed one. Use the <Pauli matrices> representation
$$
W_F=\frac14\left(I\otimes I-w\sum_{j=1}^3\sigma_j\otimes\sigma_j\right),
\qquad w=\frac{4F-1}{3}\in(1/3,1].
$$
Write the positive matrices $X=A^\dagger A=a_0I+a\cdot\sigma$ and $Y=B^\dagger B=b_0I+b\cdot\sigma$, so $a_0\geq|a|$, $b_0\geq|b|$. Taking traces gives
$$
p=a_0b_0-w\,a\cdot b
\geq a_0b_0-|a||b|
\geq\sqrt{(a_0^2-|a|^2)(b_0^2-|b|^2)}
=|\det A\det B|.
$$
For completeness, squaring the middle comparison leaves $(a_0|b|-b_0|a|)^2\geq0$, and both sides are nonnegative. This proves the <single-copy Werner filtering normalization bound>, and hence
$$
\boxed{C(\rho')\leq C(W_F)\quad\text{for every nonzero-probability branch}.}
$$
Forgetting some record information or grouping several successful records mixes these branch outputs; convexity preserves the same bound. Any alleged final $W_{F'}$ would therefore obey $2F'-1\leq2F-1$, contradicting $F'>F$. Thus \b[even heralded single-copy <LOCC> cannot produce the specified improved <Werner state>]. This proves that <single-copy Werner filtering cannot increase concurrence>. It does not forbid collective <entanglement distillation> from several noisy pairs; allowing additional shared entangled input copies would change the resource assumption.
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