contrageometric proportion

Just as one converts the proportion equation

$$\frac{m-x}{y-m}=\frac{x}{y}$$

defining the harmonic mean of $x$ and $y$ into the proportion equation

$$\frac{m-x}{y-m}=\frac{y}{x}$$

defining their contraharmonicmean (http://planetmath.org/ContraharmonicProportion), one also may convert the proportionequation

$$\frac{m-x}{y-m}=\frac{m}{y}$$

defining the geometric mean into a new equation

${\displaystyle \frac{m-x}{y-m}}={\displaystyle \frac{y}{m}}$(1)

defining the contrageometric mean $m$ of $x$ and $y$.  Thus, thethree positive numbers $x$, $m$, $y$ satisfying (1) are in contrageometric proportion.  One integer example is$1,\mathrm{\hspace{0.17em}4},\mathrm{\hspace{0.17em}6}$.

Solving $m$ from (1) one gets the expression

$m={\displaystyle \frac{x-y+\sqrt{{(x-y)}^2+4y^2}}{2}}=:f(x,y).$(2)

Suppose now that  $0\le x\le y$.  Using (2) we see that

$$m\ge \frac{x-y+\sqrt{0^2+4y^2}}{2}=\frac{x+y}{2}\ge x,$$

$$y^2-m^2=\frac{-{(y-x)}^2+(y-x)\sqrt{{(x-y)}^2+4y^2}}{2}=\frac{(y-x)[\sqrt{{(y-x)}^2+4y^2}-(y-x)]}{2}\ge \mathrm{\hspace{0.33em}0},$$

accordingly

$x\le f(x,y)\le y.$(3)

Thus the contrageometric mean of $x$ and $y$ also is at least equal totheir arithmetic mean.  We can also compare $m$ with their quadratic mean by watching the difference

$${\left(\sqrt{\frac{x^2+y^2}{2}}\right)}^2-m^2=(y-x)\left(\frac{1}{2}\sqrt{{(y-x)}^2+4y^2}-y\right)\ge \mathrm{\hspace{0.33em}0}.$$

So we have

${\displaystyle \frac{x+y}{2}}\le f(x,y)\le \sqrt{{\displaystyle \frac{x^2+y^2}{2}}}.$(4)

Cf. this result with the comparison of Pythagorean means (http://planetmath.org/ComparisonOfPythagoreanMeans); there the brown curve is the graph of  $f(x,\mathrm{\hspace{0.17em}1})$.

It’s clear that the contrageometric mean (2) is not symmetric withrespect to the variables $x$ and $y$, contrary to the other types ofmeans in general.  On the other hand, the contrageometric mean is, as other types of means, a first-degree homogeneous function its arguments:

$f(tx,ty)=tf(x,y).$(5)