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By the Schmidt decomposition theorem, suitable local unitary operations put the normalized pure state into the formIt is entangled precisely when both Schmidt coefficients are nonzero. Put $s=2{\lambda }_0{\lambda }_1$, so $0<s\le 1$ for the case of interest. The Pauli operatorssatisfy $⟨Z\otimes Z⟩=1$ and $⟨X\otimes X⟩=s$ in this pure state: $Z\otimes Z$ fixes both basis terms, and $X\otimes X$ exchanges them.

Choose Alice's two observables as $A_0=Z$, $A_1=X$, and Bob's asThese are valid dichotomic projective measurements. Indeed, $XZ+ZX=0$ and $X^2=Z^2=I$, so $B_0^2=B_1^2=I$; each observable has outcomes $\pm 1$. For the CHSH inequality convention $|⟨A_0\otimes (B_0+B_1)+A_1\otimes (B_0-B_1)⟩|\le 2$, the quantum value isFor the original pure state, conjugate these four observables by the corresponding local unitary operators. Their outcomes and Pauli correlators are unchanged, and each party still measures only their own qubit. Thus every entangled pure two-qubit state violates CHSH for suitable local measurements, proving Gisin's theorem directly. The CHSH axes for an entangled pure two-qubit state approach a nonviolating configuration continuously as one Schmidt coefficient tends to zero; the strict violation therefore need not be large for weak entanglement.

Apply the Schmidt decomposition to the bipartite pure state. Local unitary operators put it in the form $|{\psi }_{\theta }⟩=\cos \theta |00⟩+\sin \theta |11⟩$, with $0<\theta \le \pi /4$ after interchanging basis vectors. Both Schmidt coefficients are nonzero because the state is entangled. Put $s=\sin 2\theta >0$.

For this state, direct application of the Pauli operators givesFor example ${\sigma }_x\otimes {\sigma }_x$ interchanges $|00⟩$ and $|11⟩$, giving $2\cos \theta \sin \theta$; the crossed products take these two kets into their orthogonal complement.

Choose the CHSH axes for an entangled pure two-qubit state:Their associated Bloch vectors have unit length, so each is a valid two-outcome projective measurement with projectors $(I\pm A_i)/2$ or $(I\pm B_j)/2$. The correlation combination isA local hidden-variable assignment with outcomes $\pm 1$ gives absolute value at most two: exactly one of $b_0+b_1$ and $b_0-b_1$ vanishes, while the other is $\pm 2$. Averaging these assignments preserves that bound, which is the CHSH inequality.

Therefore $\boxed{\mathcal{S}=2\sqrt{1+{\sin }^{2}2\theta }>2}$. For the original state $(U\otimes V)|{\psi }_{\theta }⟩$, conjugate Alice's observables by $U$ and Bob's by $V$. These conjugations rotate their measurement axes and preserve all the displayed correlations. Thus every entangled pure two-qubit state has the requested violating axes. The maximally entangled case gives $2\sqrt{2}$, while the unentangled limit merely saturates two.

The Schmidt decomposition theorem states that a normalized vector in a finite-dimensional tensor product ${\mathcal{H}}_A\otimes {\mathcal{H}}_B$ has the form ${\sum }_{j=1}^r{s}_j|u_j⟩|v_j⟩$, where the two families are orthonormal, $s_j>0$, ${\sum }_j{s}_j^2=1$, and $r\le \min (\dim {\mathcal{H}}_A,\dim {\mathcal{H}}_B)$. Extending each family to an orthonormal basis for two qubits givesThe nonzero Schmidt coefficients can be made positive by absorbing phases into the basis vectors. Both are positive exactly when the pure state is entangled. A product state has Schmidt rank one, so one coefficient is zero; the source's assertion of two positive coefficients for every pure state needs this exception.

Choose each local orthonormal basis independently as the computational basis. In that basis ${\sigma }_z=|0⟩⟨0|-|1⟩⟨1|$. This is a local change of coordinates, implemented by separate unitary matrices, rather than a physical restriction on the original quantum state. The associated Pauli operators supply the other two local axes. Up to an irrelevant common phase, each unitary change of qubit basis corresponds to a rotation of its Bloch sphere.

Put $s=2{\lambda }_0{\lambda }_1$. The Schmidt-basis Pauli correlation tensor is diagonal. The Pauli operators ${\sigma }_x\otimes {\sigma }_x$ exchange $|00⟩$ and $|11⟩$, while ${\sigma }_y\otimes {\sigma }_y$ do so with minus signs, and ${\sigma }_z\otimes {\sigma }_z$ leaves both fixed. HenceMixed components vanish: those containing one $z$ and one transverse Pauli operator map the occupied basis vectors outside their span, while the $xy$ and $yx$ matrix elements are purely imaginary and cancel for real Schmidt coefficients. By bilinearity, for arbitrary real vectors,For a CHSH inequality test take unit quantum measurement axesThese CHSH axes for an entangled pure two-qubit state giveThe local bound is two: for each hidden state, $A(B+B^{\prime })+A^{\prime }(B-B^{\prime })$ is $\pm 2$ when all four outcomes are $\pm 1$, and averaging cannot increase its absolute value. Thus every entangled pure two-qubit state violates a CHSH inequality, the content of Gisin's theorem. A product state has $s=0$ and does not violate it, so the unqualified final claim in the question is false for that case. The maximum $2\sqrt{2}$ occurs for equal Schmidt coefficients. This excludes local hidden-variable theories satisfying measurement independence, but does not permit faster-than-light signalling: the local reduced density matrix and its quantum measurement probabilities are unchanged by the remote choice of axis.