An old question, but it got me thinking. And I did get an answer, so here it is...
We have a circuit, with $n$ nodes, numbered $0\ldots n-1$. There are also $b$ branches, numbered $0\ldots b-1$. Node number $0$ ($n_0$) is chosen to be the reference, or ground node. Node number $1$ ($n_1$) is kept $1 \mathrm{V}$ by a voltage source connected from $n_0$ to $n_1$. This is branch $0$ ($b_0$). The current on $b_0$ is defined to be flowing from $n_0$ to $n_1$, making $v_0$ (voltage of branch $0$) $-1\mathrm{V}$ by convention.
The equivalent resistance of the circuit as seen from nodes $n_1 - n_0$ is defined to be $1V / i_0$.
Now, in addition to $b_0$ consider another branch $b_1$, which is initially "not present", that is, initially $i_1 = 0$.
Here, we invoke <PERSON>'s theorem, which states the following:
Let $I_k$ be a set of branch currents on a circuit satisfying KCL.
Let $V_k$ be a set of branch voltages on a circuit satisfying KVL.
Then, for any such sets,
$\sum I_k\cdot V_k = 0$
Note that voltages and currents need only satisfy KCL and KVL, they need not be related to each other!
Now, going back to our setup, we have two sets of currents and voltages.
First, with $b_1$ not present.
$i_k$, $v_k$
$\sum i_k \cdot v_k = 0$
Expanding the special branches we have:
$v_0 i_0 + v_1 i_1 + \sum\limits_{k=2}^{b-1} v_k i_k = 0$
Since $v_0 = -1$ and $i_1 = 0$ we have:
$i_0 = \sum\limits_{k=2}^{b-1} v_k i_k$
When $b_1$ is inserted with a finite resistor, all currents and voltages are different. 628
So we get:
$i'0 = v'_1 i'_1 + \sum\limits{k=2}^{b-1} v'_k i'_k$
Unfortunately, this takes us nowhere. But, there is a different approach: Use <PERSON>'s theorem, but with voltages and currents crossed between the two situations. This is perfectly valid:
$i_0 = \sum\limits_{k=2}^{b-1} v'_k i_k$
$i'0 = v_1 i'_1 + \sum\limits{k=2}^{b-1} v_k i'_k$
Up to here, everything is general. Now, we use the fact that, apart from $b_0$ which contains a voltage source, every element is a resistor:
$v_k = R_k \cdot i_k$
and
$v'_k = R_k \cdot i'_k$
Substituting theses for the voltages, the equations become:
$i_0 = \sum\limits_{k=2}^{b-1} R_k i'_k i_k$
$i'0 = v_1 i'_1 + \sum\limits{k=2}^{b-1} R_k i_k i'_k$
Note that the sums are exactly the same! Subtract the two equations to obtain:
$i'_0 - i_0 = v_1 i'_1$
Now, $v_1$ is the voltage where we are going to connect the new resistor before we have connected it. $i'_1$ is the current going through it after we have connected it. The product is definitely non-negative. This can be proven by considering the rest of the circuit as seen by the new resistor as a <PERSON> equivalent circuit; the later current must be in the same direction as the initial voltage. 628