Answer
The geometric mean is greater or equal to the harmonic mean
Work Step by Step
Let $x$ and $y$ be two real numbers, $x>0,y>0$.
The geometric mean is:
$M_g=\sqrt{xy}$,
while the harmonic mean is:
$M_h=\dfrac{2}{\frac{1}{x}+\frac{1}{y}}=\dfrac{2xy}{x+y}$
We have:
$(x-y)^2\geq 0$
$x^2-2xy+y^2\geq 0$
$x^2-2xy+y^2+4xy\geq 0+4xy$
$x^2+2xy+y^2\geq 4xy$
$(x+y)^2\geq 4xy$
$xy(x+y)^2\geq xy(4xy)$
$xy(x+y)^2\geq 4x^2y^2$
$\dfrac{xy(x+y)^2}{(x+y)^2}\geq \dfrac{4x^2y^2}{(x+y)^2}$
$xy\geq \left(\dfrac{2xy}{x+y}\right)^2$
$\sqrt{xy}\geq \dfrac{2xy}{x+y}$
$M_g\geq M_h$
Therefore the geometric mean is greater or equal to the harmonic mean.
