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Let $d=d_G(a,b)>0$, and for $i=1,\dots ,d$ putEach edge cutset ${\Pi }_i$ separates $a$ from $b$. Adjacent vertices have graph distances from $a$ differing by at most one. Hence an edge can cross at most one of these level boundaries, so the edge cutsets are pairwise edge-disjoint. Write $k_i=|{\Pi }_i|\ge 1$; then ${\sum }_i{k}_i\le |E|$.

The Nash-Williams inequality for unit conductances giveswhere the middle inequality is the Cauchy-Schwarz inequality. One can also derive the first inequality directly: every unit flow has signed net flux one through each ${\Pi }_i$, so its energy of a flow on that cut is at least $1/k_i$; sum over the disjoint cuts and use the Thomson principle.

Combining with the commute time identity yieldsThe case $a=b$ is trivial. A path graph, with $a,b$ its endpoints, attains equality.

Thomson principle states that the effective resistance is the minimum energy of a unit flow from $a$ to $z$:An edge cutset separating $a$ and $z$ is a set of edges whose removal disconnects them. Every unit flow has net flux one across each such cutset ${\Pi }_k$. By Cauchy-Schwarz inequality,For disjoint cutsets, summing these energy lower bounds and applying Thomson's principle gives the Nash-Williams inequalityThe commute time identity is

There are $\asymp n^3$ edges. The $n^2$ horizontal cutsets between successive rows perpendicular to the long direction are disjoint and each contains $\asymp n$ unit-conductance edges. Nash-Williams inequality givesConversely, spread a unit flow nearly uniformly across the width $n$ while moving it through the $n^2$ long-direction layers, with bounded extra energy to fan out from and collect at the two corner vertices. Its energy is $O(n^2/n)+O(\log n)=O(n)$, so Thomson's principle gives the matching upper bound $R_{\mathrm{eff}}\asymp n$.

The graph is invariant under a half-turn exchanging the two corners, so the two directional hitting-time expectations are equal. The commute time identity therefore yields