How to prove that if $x>0$ and $y>0$, then $\sqrt{x}+\sqrt{y}>\sqrt{x+y}$, using the relation of arithmetic and geometric means? 
How to prove that if $x>0$ and $y>0$, then $$\sqrt{x}+\sqrt{y}>\sqrt{x+y}\,,$$ using the relation of arithmetic and geometric means?

I started by showing that if $x>0$ and $y>0$, based on the relation of arithmetic and geometric means, $\dfrac{x+y}{2}\ge\sqrt{xy}$.
Hence, $x+y\ge2\sqrt{xy}$.
I am now stuck here and don't know what must be the next step.
Any suggestions or comments will be much appreciated.

Is this correct?
Since $x+y≥2\sqrt{xy}$, by AM-GM relationship,
$x+y+2\sqrt{xy}≥x+y$
$(\sqrt{x}+\sqrt{y})^2≥x+y$
Thus, $\sqrt{x}+\sqrt{y}>\sqrt{x+y}$. QED.
