[tex]\large\rm \dfrac{x+y}{2}\ge \sqrt{xy}[/tex]
Squaring each side gives us,
[tex]\large\rm \dfrac{(x+y)^2}{4}\ge xy[/tex]
Multiply by 4,
[tex]\large\rm (x+y)^2\ge 4xy[/tex]
and subtract,
[tex]\large\rm (x+y)^2-4xy\ge 0[/tex]
Expand out the square,
[tex]\large\rm x^2+2xy+y^2-4xy\ge0[/tex]
Combine like-terms,
[tex]\large\rm x^2-2xy+y^2\ge0[/tex]
This can now be written as a different perfect square,
[tex]\large\rm (x-y)^2\ge0[/tex]
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.
