# Where did these classical means come from?
For two positive numbers $x,y > 0$, we are familiar with some ways to calculate their "means", for instance their **arithmetic mean** $$
a(x,y) = \frac{1}{2}(x+y),
$$their **geometric mean** $$
g(x,y) = \sqrt{xy},
$$and their **harmonic mean** $$
h(x,y) = \frac{2}{\frac{1}{x}+\frac{1}{y}}.
$$But how were these dreamt up, and are there others?
These means arise from antiquity Greeks, and to capture this notion of mean of two positive numbers, they looked at **ratios**. Take $0 < x < y$, and say $m$ is some kind of mean for both numbers, satisfying $x < m < y$. Then we can consider the ratios of the two segments $y-m$ to $m-x$, and set them equal to the nine possible ratios you can make with $x,y,m$. That is, we can assign $$
\frac{y-m}{m-x}
$$ to one of the nine ratios $$\frac{y-m}{m-x}= \,\,\,
\begin{array}{}\displaystyle
\frac{x}{x}, & \displaystyle\frac{x}{m}, & \displaystyle\frac{x}{y}, \\
\displaystyle\frac{m}{x}, & \displaystyle\frac{m}{m}, & \displaystyle\frac{m}{y}, \\
\displaystyle\frac{y}{x}, & \displaystyle\frac{y}{m}, & \displaystyle\frac{y}{y}.
\end{array}
$$
## Case analyses.
Let us analyze all 9 cases, and see what kind of means we get.
### Cases $x/x$, $m/ m$, and $y /y$.
Note if we set $$
\frac{y-m}{m-x} = \frac{x}{x} = \frac{m}{m} = \frac{y}{y} = 1,
$$then $2m = x+y$, hence we get $$
m = \frac{1}{2}(x+y) = a(x,y)
$$the arithmetic mean of $x,y$.
### Case $x / m$.
If we set $$
\frac{y-m}{m-x} = \frac{x}{m}
$$then we get $my - m^{2} = xm - x^{2}$, then $m^{2} + (x-y)m - x^{2} =0$, so $$
m = \frac{y-x + \sqrt{(y-x)^{2}+4x^{2}}}{2}
$$where we have to have the $+$ sign for $m > 0$. We don't have a conventional name for this mean. Let us denote it $r(x,y)$.
### Case $x / y$.
If we set $$
\frac{y-m}{m-x} = \frac{x}{y}
$$then we get $y^{2}-my = mx - x^{2}$, so $x^{2}+y^{2}=m(x+y)$, so $$
m = \frac{x^{2}+y^{2}}{x+y}
$$
We don't have a conventional name for this. Let us denote it $t(x,y)$.
### Case $m / x$ and $y / m$.
If we set $$
\frac{y-m}{m-x} = \frac{m}{x}
$$then we get $xy - mx = m^{2}-mx$, so $$
m = \sqrt{xy},
$$which is the geometric mean of $x,y$.
Similarly, if we have $$
\frac{y-m}{m-x} = \frac{y}{m}
$$then we get $my - m^{2} = my-xy$, so $m = \sqrt{xy}$ the geometric mean.
### Case $m / y$.
If we set $$
\frac{y-m}{m-x} = \frac{m}{y}
$$then we get $y^{2}-my = m^{2}-mx$, so $m^{2} + (y-x)m - y^{2} = 0$, so $$
m = \frac{-(y-x) + \sqrt{(y-x)^{2}+4y^{2}}}{2}.
$$We don't have a conventional name for this mean. Let us denote it $s(x,y)$.
### Case $y / x$.
If we set $$
\frac{y-m}{m-x} = \frac{y}{x}
$$then we get $xy - mx = my - xy$, or $2xy = m(x+y)$, or $$
m = \frac{2xy}{x+y} = \frac{2}{\frac{1}{x}+\frac{1}{y}} = h(x,y)
$$which is the harmonic mean of $x,y$.
So the means $m$ that we get by setting $\displaystyle \frac{y-m}{m-x} = \frac{p}{q}$ where $p,q \in \{x,m,y\}$, for $0 < x < m < y$, gives the following table $$
\begin{array}{cc|ccc}
& & & q \\
& & x & m & y \\ \hline
& x & a & r & t \\
p & m & g & a & s \\
& y & h & g & a
\end{array}
$$where $a(x,y) = \frac{1}{2} (x+y)$, $g(x,y) = \sqrt{xy}$, $h(x,y) = \frac{2}{\frac{1}{x}+\frac{1}{y}}$, $r(x,y) = \frac{y-x + \sqrt{(y-x)^{2}+4x^{2}}}{2}$, $s(x,y) = \frac{-(y-x) + \sqrt{(y-x)^{2}+4y^{2}}}{2}$, and $t(x,y) = \frac{x^{2} +y^{2}}{x+y}$.
Observe that $r(x,y)$ and $s(x,y)$ are not symmetric in the arguments, while the rest are symmetric functions. All these means are homogeneous, however, that for all $\lambda > 0$, $$
f(\lambda x, \lambda y) = \lambda f(x,y)
$$where $f \in \{t, r, s, a, g, h\}$, $0 < x < y$.
## Inequalities.
> **Claim / Conjecture** We claim we have for $0 < x< y$, $$
\max(x,y) \ge t(x,y) \ge r(x,y) \ge s(x,y) \ge a(x,y) \ge g(x,y) \ge h(x,y)\ge \min(x,y).
$$
For the means $m$ equal to $t$, $r$, or $s$, note their constructions are such that $$
\frac{y-m}{m-x} < 1 \implies m > \frac{x + y}{2} = a(x,y)
$$so we get $t,r,s > a$.
And for $0 < x < m < y$, the ratios $$
\frac{x}{m} > \frac{x}{y} \text{ and } \frac{m}{y} > \frac{x}{y}
$$so $r < t$ and $s < t$ respectively.
**Remark.** By this same vein, it is also very easy to show $a > g > h$, by using their ratio definitions!
### What about $r$ and $s$ ?
It remains to show $r > s$
Since these means are all homogeneous, it suffices to show for $x > 1$, $$
r(1,x) \ge s(1, x)
$$
Note $$
r(1,x) = \frac{x-1 + \sqrt{(x-1)^{2}+4}}{2}
$$and $$
s(1,x) = \frac{-(x-1) + \sqrt{(x-1)^{2}+4x^{2}}}{2}
$$
Note $$
\begin{align*}
r(1,x) - s(1,x) & = x-1 + \frac{1}{2}(\sqrt{(x-1)^{2}+4}-\sqrt{(x-1)^{2}+4x^{2}}) \\
\end{align*}
$$
Is there a nice way to see that this is $\ge 0$ for all $x > 1$?