order of six means

The of the six usual means of two positive numbers ($a$ and $b$) is from the least to the greatest one

1. harmonic mean,

2. geometric mean,

3. Heronian mean,

4. arithmetic mean,

5. quadratic mean,

6. contraharmonic mean,

i. e.

$$\frac{2ab}{a+b}\leqq \sqrt{ab}\leqq \frac{a+\sqrt{ab}+b}{3}\leqq \frac{a+b}{2}\leqq \sqrt{\frac{a^2+b^2}{2}}\leqq \frac{a^2+b^2}{a+b}.$$

The equality signs are valid iff  $a=b$.

Proof.  If  $x^2-y^2\geqq 0$  for nonnegative $x$ and $y$,  then  $x\geqq y$.

“$1\leqq 2$”:
${\left(\sqrt{ab}\right)}^2-{\left({\displaystyle \frac{a+b}{2}}\right)}^2=ab-{\displaystyle \frac{4a^2{b}^2}{{(a+b)}^2}}=ab\left(1-{\displaystyle \frac{4ab}{{(a+b)}^2}}\right)=ab\cdot {\displaystyle \frac{{(a+b)}^2-4ab}{{(a+b)}^2}}={\displaystyle \frac{ab{(a-b)}^2}{{(a+b)}^2}}\geqq 0$

“$2\leqq 3$” and “$3\leqq 4$”: proven in  Heronian mean is between geometric and arithmetic mean

“$4\leqq 5$”:
${\left(\sqrt{{\displaystyle \frac{a^2+b^2}{2}}}\right)}^2-{\left({\displaystyle \frac{a+b}{2}}\right)}^2={\displaystyle \frac{2a^2+2b^2-a^2-2ab-b^2}{4}}={\left({\displaystyle \frac{a-b}{2}}\right)}^2\geqq 0$

“$5\leqq 6$”:
${\left({\displaystyle \frac{a^2+b^2}{a+b}}\right)}^2-{\left(\sqrt{{\displaystyle \frac{a^2+b^2}{2}}}\right)}^2={\displaystyle \frac{2{(a^2+b^2)}^2-(a^2+b^2){(a+b)}^2}{2{(a+b)}^2}}={\displaystyle \frac{(a^2+b^2)(2a^2+2b^2-a^2-2ab-b^2)}{2{(a+b)}^2}}={\displaystyle \frac{(a^2+b^2){(a-b)}^2}{2{(a+b)}^2}}\geqq 0$

Title	order of six means
Canonical name	OrderOfSixMeans
Date of creation	2013-03-22 18:45:28
Last modified on	2013-03-22 18:45:28
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	4
Author	pahio (2872)
Entry type	Theorem
Classification	msc 06A05
Classification	msc 26B35
Classification	msc 26D07
Related topic	Mean3
Related topic	ComparisonOfPythagoreanMeans
Related topic	InequalityWithAbsoluteValues
Related topic	LehmerMean