Equivalent resistance of a set of n resistors?
Let's see if this works.

Start with \$n=2\$:

$$
R_s = R_1+R_2\\
R_p = \frac{R_1R_2}{R_1+R_2}\\
\text{Then: }\\
\frac{R_1R_2}{R_1+R_2} \stackrel{?}{&lt;}R_1+R_2\\
R_1R_2 \stackrel{?}{&lt;} (R_1+R_2)^2
$$
The last row can be derived from the one before if \$R_1+R_2&gt;0\$ and it is true if both resistance are positive separately too, which is something ok to ask to a resistance. \$n=2\4 is our base case.

Now... induction.

Hypothesis: \$R_p[n] &lt; R_s[n]\$

Thesis: \$R_p[n+1] &lt; R_s[n+1]\$

We try to prove that the sequence \$R_p[n]\$ is decreasing, while \$R_s[n]\$ is increasing.

Series resistance:
$$
R_s[n+1] = R_s[n] + R_{n+1} &gt; R_s[n] \text{ if } R_{n+1}&gt;0\\
R_s[n+1] &gt; R_s[n] \textit{ q.e.d.}
$$

Parallel resistance:
$$
R_p[n+1] = \frac{R_p[n]R_{n+1}}{R_p[n]+R_{n+1}}\stackrel{?}{&lt;}R_p[n]\\
R_p[n]R_{n+1}\stackrel{?}{&lt;}R_p[n]^2+R_p[n]R_{n+1}\\
R_p[n]^2&gt;0\text{ which is true, so that}
R_p[n+1] &lt; R_p[n] \textit{ q.e.d.}
$$

We finally chain things up:

$$
R_p[n+1] &lt; R_p[n] &lt; R_s[n] &lt; R_s[n+1]
$$

It did work after all.