Heronian mean is between geometric and arithmetic mean

Theorem.  For non-negative numbers $x$ and $y$, the inequalities

$$\sqrt{xy}\leqq \frac{x+\sqrt{xy}+y}{3}\leqq \frac{x+y}{2}$$

are in , i.e. the Heronian mean is always at least equal to the geometric mean and at most equal to the arithmetic mean.  The equality signs are true if and only if  $x=y$.

Proof.
${\mathrm{1}}^{\mathrm{\circ }}\mathrm{.}$

$\sqrt{xy}\leqq {\displaystyle \frac{x+\sqrt{xy}+y}{3}}$ $\mathrm{\hspace{1em}}\iff \mathit{\hspace{1em}}3\sqrt{xy}\leqq x+\sqrt{xy}+y$

$\mathrm{\hspace{1em}}\iff \mathit{\hspace{1em}}2\sqrt{xy}\leqq x+y$

$\mathrm{\hspace{1em}}\iff \mathit{\hspace{1em}}4xy\leqq x^2+2xy+y^2$

$\mathrm{\hspace{1em}}\iff \mathit{\hspace{1em}}0\leqq x^2-2xy+y^2$

$\mathrm{\hspace{1em}}\iff \mathit{\hspace{1em}}0\leqq {(x-y)}^2$

$2^{\circ }.$

${\displaystyle \frac{x+\sqrt{xy}+y}{3}}\leqq {\displaystyle \frac{x+y}{2}}$ $\mathrm{\hspace{1em}}\iff \mathit{\hspace{1em}}2x+2\sqrt{xy}+2y\leqq 3x+3y$

$\mathrm{\hspace{1em}}\iff \mathit{\hspace{1em}}2\sqrt{xy}\leqq x+y$

$\mathrm{\hspace{1em}}\iff \mathit{\hspace{1em}}4xy\leqq x^2+2xy+y^2$

$\mathrm{\hspace{1em}}\iff \mathit{\hspace{1em}}0\leqq {(x-y)}^2$

All inequalities of both chains are equivalent (http://planetmath.org/Equivalent3) since $x$ and $y$ are non-negative.  As for the equalities, the chains are valid with the mere equality signs.

Title	Heronian mean is between geometric and arithmetic mean
Canonical name	HeronianMeanIsBetweenGeometricAndArithmeticMean
Date of creation	2013-03-22 17:49:14
Last modified on	2013-03-22 17:49:14
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	10
Author	pahio (2872)
Entry type	Theorem
Classification	msc 26B99
Classification	msc 26D07
Classification	msc 01A20
Classification	msc 00A05
Synonym	Heronian mean inequalities
Related topic	ArithmeticGeometricMeansInequality
Related topic	ComparisonOfPythagoreanMeans
Related topic	SquareOfSum
Related topic	Equivalent3
Related topic	HeronsPrinciple