= Solution
Apply the <OSSS inequality> to the independent coordinates $(V_x,W_x)$ and to the indicator of the <one-arm event> $A_R(a)$. Use the randomized <OSSS exploration of a one-arm event>: choose $k$ uniformly from $\{1,\ldots,R\}$ and reveal the variables needed to explore the superlevel cluster meeting $\partial B_k$. A coordinate can be revealed only if a nearby vertex has an open connection over the relevant distance. Translation invariance, finite-range dependence and the <union bound> therefore give the revealment estimate
$$
\delta_x\leq\frac{C_d}{R}\sum_{k=1}^R\theta_k
$$
for both kinds of coordinates, after enlarging the explored neighbourhood by a distance depending only on $d$.
For an increasing Gaussian threshold event, the resampling influence of $V_x$ is bounded by a universal constant times $\mathbb E[V_x\mathbf1_{A_R(a)}]$. The influence of $W_x$ is bounded by the sum of the corresponding $V$ influences at the $2d$ neighbours of $x$: indeed <Gaussian integration by parts> gives
$$
\mathbb E[W_x\mathbf1_{A_R(a)}]
=\frac1{2d}\sum_{z\sim x}\mathbb E[V_z\mathbf1_{A_R(a)}],
$$
first for smooth increasing approximations and then by a limit. Consequently the OSSS bound becomes
$$
\theta_R(1-\theta_R)
\leq C_d\left(\frac1R\sum_{k=1}^R\theta_k\right)
\sum_{x\in B_R}\mathbb E[V_x\mathbf1_{A_R(a)}].
$$
Part (c) identifies the final sum with $-\theta_R'(a)$. Dividing and putting $c=C_d^{-1}>0$ proves
$$
-\frac d{da}\theta_R
\geq c\,
\frac{\theta_R(1-\theta_R)}{R^{-1}\sum_{k=1}^R\theta_k}.
$$
Solved by gpt-5.6-sol high.
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