dropping the $o(n)$-term is of no consequence in the following computations. This implies that if $t$ is such that $|E(\sigma_t,\overline{\sigma_t})|
-|E(\sigma,\overline{\sigma})|$ is maximised, then we get (again, possibly after dropping a term of order $o(n)$) $$\begin{aligned}
\sum_{i=1}^t \big(\left|E\left(v_{i},\sigma_{i-1}\right)\right|
-\left|E\left(v_{i},\overline{\sigma_{i-1}}\right)\right|\big)
& = & \sum_{i=t+1}^s \big(\left|E\left(v_{i},\overline{\sigma_{i-1}}\right)\right|
-\left|E\left(v_{i},\sigma_{i-1}\right)\right|\big).
\label{eq:peakequality}
\end{aligned}$$ Let $m_{t}$ denote the left-hand side of $\eqref{eq:peakequality}$, so that $$\begin{aligned}
m_{t} &= \sum_{i=1}^{t}\left(d_{i}-2\left|E\left(v_{i},\overline{\sigma_{i-1}}\right)\right|\right)
=  \sum_{i=t+1}^{s}\left(d_{i}-2\left|E\left(v_{i},\sigma_{i-1}\right)\right|\right)\\
&=  \sum_{i=1}^{s}\left(d_{i}-2\left|E\left(v_{i},\sigma_{i-1}\right)\right|\right)
-\sum_{i=1}^{t}\left(d_{i}-2\left|E\left(v_{i},\sigma_{i-1}\right)\right|\right).
\end{aligned}$$ Hence $$\sum_{i=1}^{t}d_{i}\left(\frac{\sum_{i=1}^{t}\left(d_{i}
-2\left|E\left(v_{i},\overline{\sigma_{i-1}}\right)\right|\right)}{\sum_{i=1}^{t}d_{i}}+1\right)
=\sum_{i=1}^{s}d_{i}-\sum_{i=t+1}^{s}2\left|E\left(v_{i},\sigma_{i-1}\right)\right|,$$ and thus $$\begin{aligned}
m_{t} &=  \left(\frac{\sum_{i=1}^{t}\left(d_{i}
-2\left|E\left(v_{i},\overline{\sigma_{i-1}}\right)\right|\right)}
{\sum_{i=1}^{t}d_{i}}\right)\left(\frac{\sum_{i=1}^{t}\left(d_{i}
-2\left|E\left(v_{i},\overline{\sigma_{i-1}}\right)\right|\right)}
{\sum_{i=1}^{t}d_{i}}+1\right)^{-1}\\
&\qquad\qquad\qquad \times \left(\sum_{i=1}^{s}d_{i}
-\sum_{i=t+1}^{s}2\left|E\left(v_{i},\sigma_{i-1}\right)\right|\right)\\
&\leq \frac{1}{2}\left(1-2\frac{I_{\delta}\left(x\right)}{x}\right)
\left(1-\frac{I_{\delta}\left(x\right)}{x}\right)^{-1}\left(x\ell_{n}
-2I_{\delta}\left(x\right)\ell_{n}+\frac{h\left(\left|\sigma\right|-s\right)}{J}\right),
\end{aligned}$$ where for the last inequality we use – and the monotonicity of $y\rightarrow y(y+1)^{-1}$. From , using the fact that $nd_{\mathrm{ave}}=\ell_{n}+o\left(n\right)$, we get that $ \mathcal{H}\left(\sigma_{t}\right) - \mathcal{H}\left(\sigma\right) < \Gamma^{\star}$ whenever $$\frac{h}{J\ell_{n}}\left(2\tilde{m}+t+\left(\left|\sigma\right|
-s\right)\left(\frac{x-2I_{\delta}\left(x\right)}{x-I_{\delta}\left(x\right)}\right)\right)
<2I_{d_{\mathrm{ave}}}\left(\tfrac{1}{2}\right)
-\left(x-2I_{\delta}\left(x\right)\right)^{2}\left(x-I_{\delta}\left(x\right)\right)^{-1}.
\label{eq:ihalfbound}$$ Note that if $x<2I_{\delta}\left(x\right)$, then for sufficiently small $h$ we can find a monotone downhill path to $\boxminus$. More precisely, $x<2I_{\delta}\left(x\right)$ implies that the terms in the right-hand side of become negative, and hence for $\frac{h}{J}<d_{\min}(\frac{2I_{\delta}\left(x\right)}{x}-1)$ every step in our path is a downhill step. For $x\geq2I_{\delta}\left(x\right)$, observe first that since the function $u\rightarrow\frac{\left(1-2u\right)^{2}}{\left(1-u\right)}$ is non-increasing for $u\leq\frac{1}{2}$, by Lemma \[lem:Ideltamonotone\] $$\frac{\left(x-2I_{\delta}\left(x\right)\right)^{2}}{\left(x-I_{\delta}\left(x\right)\right)}
= x\frac{(1-2\frac{I_{\delta}\left(x\right)}{x})^{2}}{(1-\frac{I_{\delta}\left(x\right)}{x})}
\leq \tfrac{1}{2}\frac{(1-4I_{\delta}(\tfrac{1}{2}))^{2}}
{(1-2I_{\delta}\left(\tfrac{1}{2}\right))},$$ and thus a sufficient condition for to 