# Heronian mean is between geometric and arithmetic mean

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.
$1^{\circ}.$

 $\displaystyle\sqrt{xy}\;\leqq\;\frac{x\!+\!\sqrt{xy}\!+\!y}{3}$ $\displaystyle\quad\Leftrightarrow\quad 3\sqrt{xy}\leqq x\!+\!\sqrt{xy}\!+\!y$ $\displaystyle\quad\Leftrightarrow\quad 2\sqrt{xy}\leqq x\!+\!y$ $\displaystyle\quad\Leftrightarrow\quad 4xy\leqq x^{2}\!+\!2xy\!+\!y^{2}$ $\displaystyle\quad\Leftrightarrow\quad 0\leqq x^{2}\!-\!2xy\!+\!y^{2}$ $\displaystyle\quad\Leftrightarrow\quad 0\leqq(x\!-\!y)^{2}$

$2^{\circ}.$

 $\displaystyle\frac{x\!+\!\sqrt{xy}\!+\!y}{3}\leqq\frac{x\!+\!y}{2}$ $\displaystyle\quad\Leftrightarrow\quad 2x\!+\!2\sqrt{xy}\!+\!2y\leqq 3x\!+\!3y$ $\displaystyle\quad\Leftrightarrow\quad 2\sqrt{xy}\leqq x\!+\!y$ $\displaystyle\quad\Leftrightarrow\quad 4xy\leqq x^{2}\!+\!2xy\!+\!y^{2}$ $\displaystyle\quad\Leftrightarrow\quad 0\leqq(x\!-\!y)^{2}$

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