# 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

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  $x\geqq y$.

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

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

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

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

 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