Question

Explain why tle=" \sqrt{xy} \leq \frac{x+y}{2}" alt=" \sqrt{xy} \leq \frac{x+y}{2}" align="absmiddle" class="latex-formula"> for any two positive real numbers x and y, $x,y \neq 0$ (The geometric mean is always less than the arithmetic mean.)

Answer 1

$\large\rm \dfrac{x+y}{2}\ge \sqrt{xy}$

Squaring each side gives us,
$\large\rm \dfrac{(x+y)^2}{4}\ge xy$

Multiply by 4,
$\large\rm (x+y)^2\ge 4xy$

and subtract,
$\large\rm (x+y)^2-4xy\ge 0$


Expand out the square,
$\large\rm x^2+2xy+y^2-4xy\ge0$

Combine like-terms,
$\large\rm x^2-2xy+y^2\ge0$

This can now be written as a different perfect square,
$\large\rm (x-y)^2\ge0$

Recall that a squared value is always positive. So (x-y)^2 is always greater than zero.

That should "explain it" sufficiently unless you need to do some sort of formal proof.