= Solution
Before $\tau$, put $a_s=\sigma(Y_s)-\sigma(X_s)$ and $d_s=b_2(Y_s)-b_1(X_s)$, so $dZ_s=a_s\,dB_s+d_s\,ds$. For $Z_s\geq0$, the Lipschitz and drift-order hypotheses give $|a_s|\leq KZ_s$ and $d_s\geq b_1(Y_s)-b_1(X_s)\geq-LZ_s$. The derivatives of $f_\epsilon(z)=(z+\epsilon)^{-1}$ are $f'_\epsilon=-(z+\epsilon)^{-2}$ and $f''_\epsilon=2(z+\epsilon)^{-3}$. Itô's formula gives
$$
\boxed{f_\epsilon(Z_{t\wedge\tau})=f_\epsilon(z_0)-\int_0^{t\wedge\tau}\frac{a_s}{(Z_s+\epsilon)^2}\,dB_s+\int_0^{t\wedge\tau}\left(-\frac{d_s}{(Z_s+\epsilon)^2}+\frac{a_s^2}{(Z_s+\epsilon)^3}\right)ds.}
$$
The drift integrand is bounded above by
$$
\frac{LZ_s}{(Z_s+\epsilon)^2}+\frac{K^2Z_s^2}{(Z_s+\epsilon)^3}\leq\frac{L+K^2}{Z_s+\epsilon}.
$$
The stochastic integrand has magnitude at most $KZ_s/(Z_s+\epsilon)^2\leq K/(4\epsilon)$, so its integral on a finite horizon is a square-integrable <martingale> with mean zero. Global Lipschitz coefficients give finite second moments of the SDE solutions on finite horizons, which also justify the drift <expectation>; alternatively localize first and use boundedness of $f_\epsilon$ and Fatou. Since $Z_{t\wedge\tau}\geq0$, taking <expectations> and replacing the stopped integral by the larger positive full-time bound yields the <reciprocal barrier proof of scalar diffusion comparison>:
$$
\boxed{\mathbb E f_\epsilon(Z_{t\wedge\tau})\leq f_\epsilon(z_0)+(L+K^2)\int_0^t\mathbb E f_\epsilon(Z_{s\wedge\tau})\,ds.}
$$
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