= Solution
For the coordinatewise lattice operations $\omega\vee\eta$ and $\omega\wedge\eta$, <Holley's condition> is
$$
\boxed{\mu_1(\omega\vee\eta)\mu_2(\omega\wedge\eta)
\ge\mu_1(\omega)\mu_2(\eta)\quad\text{for all }\omega,\eta.}
$$
This sufficient condition implies $\mu_1\ge\mu_2$ by the <Holley inequality>. It is often stated first for strictly positive <probability> weights, as will hold for the random-cluster application, but the displayed condition is also sufficient for nonnegative weights.
For clarity about the zero-weight case, apply the <four functions theorem> to an <increasing event> $A$, with the first two functions $\mu_1\mathbf1_{A^c}$ and $\mu_2\mathbf1_A$, and the last two $\mu_1\mathbf1_A$ and $\mu_2\mathbf1_{A^c}$. If $\omega\notin A$ and $\eta\in A$, then $\omega\vee\eta\in A$ and $\omega\wedge\eta\notin A$. The pointwise hypothesis is exactly the displayed cross-lattice inequality. Its summed conclusion gives
$$
\mu_1(A^c)\mu_2(A)\le\mu_1(A)\mu_2(A^c),
$$
which rearranges to $\mu_1(A)\ge\mu_2(A)$. This establishes the same statement without a positivity-of-support assumption.
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