The AM-GM inequality says that

*harmonic mean ≤ geometric mean ≤ arithmetic mean.*

But it's not always clear

why it should be so. The

calculus aspects are well covered by the

Jensen inequality proof utilising

convexity of -

log(

`x`), or using various versions of

Lagrange multipliers. For 2 numbers, it's an

isoperimetric inequality that says that of the rectangles of given

perimeter the

square has the greatest

area. But for

`n` numbers, it's not clear what it says.

Here's a proof of just the HM<=AM bit which *makes sense* physically (and can also be made completely rigourous). Any extension to cover the GM will be **very much** welcomed by the author! I came up with it on my own. I'm unaware of anyone else's work along these lines -- please do inform me if I'm unoriginal.

Suppose we have resistors of values R_{1},...,R_{n}. Arrange them in two ways:

- The "series parallel" network
- This has n parallel columns of resistors arranged in series, so it has effective resistance equal to the harmonic mean of the resistances:
+-R1-+ +-R1-+ +-R1-+
| | | | | |
+-R2-+ +-R2-+ +-R2-+
| . | | . | | . |
--+ . +--+ . +--...--+ . +--
| . | | . | | . |
+-Rn-+ +-Rn-+ +-Rn-+

Each of the n identical parallel block has effective resistance 1/(1/R_{1}+...+1/R_{n}), so the total resistance is n times that, or the harmonic mean.
- The "parallel series" network
- This has n parallel rows of arrangements of all the resistors in series, so it has effective resistance equal to the arithmetic mean of the resistances:
+--R1--R2--...--Rn--+
| |
+--R1--R2--...--Rn--+
| |
--+ ... +--
| |
+--R1--R2--...--Rn--+

Each of n identical rows has resistance R_{1}+...+R_{n}, so the effective resistance is 1/n of the resistance of each row, or the arithmetic mean of the resistances.

We want to show the effective resistance of the first is less than or equal to that of the second. To do this we

*rearrange* each row in the "parallel series" network, so that every resistor appears once in each column:

+--R1--R2--...--Rn--+
| |
+--R2--R3--...--R1--+
| |
--+ ... +--
| |
+--Rn--R1--...-Rn-1-+

Obviously, the effective resistance of this "revised parallel series" network is unchanged by this exercise, and it looks as though we're just pushing pieces of paper around.

**BUT!** "Recall" the obviously true (although not so easy to prove) theorem of electrical networks that shorting 2 points in an electrical network cannot increase its effective resistance (in fact, it is true that the effective resistance of a network of resistors is a monotone increasing function of the resistances; here we're just using a special case that lets us change resistance ∞ to resistance 0).

Armed with this theorem (and a soldering iron), let's start proving! Run a short between every two adjacent columns of the "revised parallel series" network (i.e.: Run the first wire from the point between R_{1} and R_{2} on the first row to the point between R_{2} and R_{3} on the second row to ... to the point between R_{n} and R_{1} on the last row; this takes care of the first column. Continue running wires down the remaining columns). Since we're just shorting pairs of points, the aforementioned theorem says we never increase the effective resistance.

But at the end of the process we're left with the "series parallel" network, which has effective resistance equal to the harmonic mean of R_{1},...,R_{n}. So the harmonic mean is no greater than the arithmetic mean.

**QED.**