# 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 contraharmonic mean (http://planetmath.org/ContraharmonicProportion), one also may convert the proportion equation

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

defining the geometric mean into a new equation

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

defining the contrageometric mean $m$ of $x$ and $y$.  Thus, the three positive numbers $x$, $m$, $y$ satisfying (1) are in contrageometric proportion.  One integer example is $1,\,4,\,6$.

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

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

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

 $m\;\geq\;\frac{x\!-\!y+\sqrt{0^{2}\!+\!4y^{2}}}{2}\;=\;\frac{x\!+\!y}{2}\;\geq% \;x,$
 $y^{2}\!-\!m^{2}\;=\;\frac{-(y\!-\!x)^{2}+(y\!-\!x)\sqrt{(x\!-\!y)^{2}+4y^{2}}}% {2}\;=\;\frac{(y\!-\!x)[\sqrt{(y\!-\!x)^{2}+4y^{2}}-(y\!-\!x)]}{2}\;\geq\;0,$

accordingly

 $\displaystyle x\;\leq\;f(x,\,y)\;\leq\;y.$ (3)

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

 $\left(\sqrt{\frac{x^{2}\!+\!y^{2}}{2}}\right)^{2}-m^{2}\;=\;(y\!-\!x)\left(% \frac{1}{2}\sqrt{(y\!-\!x)^{2}+4y^{2}}-y\right)\;\geq\;0.$

So we have

 $\displaystyle\frac{x\!+\!y}{2}\;\leq\;f(x,\,y)\;\leq\;\sqrt{\frac{x^{2}\!+\!y^% {2}}{2}}.$ (4)

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

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

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

## References

• 1 Mabrouk K. Faradj: http://etd.lsu.edu/docs/available/etd-07082004-091436/unrestricted/Faradj_thesis.pdfWhat mean do you mean? An exposition on means.  Louisiana State University (2004).
• 2 Georghe Toader & Silvia Toader: http://rgmia.org/papers/monographs/Grec.pdfGreek means and the arithmetic-geometric mean.  RGMIA (2010).
Title contrageometric proportion ContrageometricProportion 2013-04-19 7:14:46 2013-04-19 7:14:46 pahio (2872) pahio (2872) 17 pahio (2872) Definition msc 26E60 msc 11-00 msc 01A20 msc 01A17 ContraharmonicProportion contrageometric mean