= Solution
Apply the <Schmidt decomposition> to the bipartite <pure state>. Local <unitary operators> put it in the form $|\psi_\theta\rangle=\cos\theta|00\rangle+\sin\theta|11\rangle$, with $0<\theta\leq\pi/4$ after interchanging basis vectors. Both <Schmidt coefficients> are nonzero because the state is <entangled>. Put $s=\sin2\theta>0$.
For this state, direct application of the <Pauli operators> gives
$$
\langle\sigma_z\otimes\sigma_z\rangle=1,\qquad
\langle\sigma_x\otimes\sigma_x\rangle=s,\qquad
\langle\sigma_z\otimes\sigma_x\rangle=\langle\sigma_x\otimes\sigma_z\rangle=0.
$$
For example $\sigma_x\otimes\sigma_x$ interchanges $|00\rangle$ and $|11\rangle$, 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>:
$$
A_0=\sigma_z,\qquad A_1=\sigma_x,\qquad
B_0=\frac{\sigma_z+s\sigma_x}{\sqrt{1+s^2}},\qquad
B_1=\frac{\sigma_z-s\sigma_x}{\sqrt{1+s^2}}.
$$
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 is
$$
\begin{aligned}
\mathcal S&=\langle A_0\otimes(B_0+B_1)+A_1\otimes(B_0-B_1)\rangle\\
&=\frac{2\langle\sigma_z\otimes\sigma_z\rangle+2s\langle\sigma_x\otimes\sigma_x\rangle}{\sqrt{1+s^2}}
=2\sqrt{1+s^2}.
\end{aligned}
$$
A local hidden-variable assignment with outcomes $\pm1$ gives absolute value at most two: exactly one of $b_0+b_1$ and $b_0-b_1$ vanishes, while the other is $\pm2$. Averaging these assignments preserves that bound, which is the <CHSH inequality>.
Therefore \b[$\boxed{\mathcal S=2\sqrt{1+\sin^22\theta}>2}$]. For the original state $(U\otimes V)|\psi_\theta\rangle$, 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 $2\sqrt2$, while the unentangled limit merely saturates two.
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