How do I prove that the inequality $\frac { 2xy }{ x+y } \le \sqrt { xy } \le \frac { x+y }{ 2 } $ holds for every $x,y>0$ Prove that for every $x,y>0$, the following inequality holds. Begin by proving the right side first and assume that $(\sqrt { x } -\sqrt { y } )^2\ge 0$
$$\frac { 2xy }{ x+y } \le \sqrt { xy } \le \frac { x+y }{ 2 } $$
What I did thus far:
Assuming: $x,y>0$ and $(\sqrt { x } -\sqrt { y } )^2\ge 0$, I will prove that $\sqrt { xy } \le \frac { x+y }{ 2 } $
1) By expanding $(\sqrt { x } -\sqrt { y } )^2\ge 0$, we get $x-2\sqrt { xy } +y\ge 0$
2) Then, $x+y\ge 2\sqrt { xy }$
3) Then, $\frac { x+y }{ 2 } \ge \sqrt { xy } $
4) Therefore, $\sqrt { xy } \le \frac { x+y }{ 2 } $
Assuming: $x,y>0$ and $(\sqrt { x } -\sqrt { y } )^2\ge 0$, I will prove that $\frac { 2xy }{ x+y } \le \sqrt { xy } $
At this point, I get stuck. I don't know what do to prove the other side of the inequality. I also don't know if I am going about this proof the right way. 
Hints are much more appreciated than the actual solution.

Edit: (added this after getting advice from answers below)
1) By expanding $(\sqrt { x } -\sqrt { y } )^2\ge 0$, we get $x-2\sqrt { xy } +y\ge 0$
2) Then, $x+y\ge 2\sqrt { xy } $
3) By taking the reciprocal of both sides, we get: $\frac { 1 }{ x+y } \le \frac { 1 }{ 2\sqrt { xy }  } $ 
4) Then, $(\sqrt { xy } )\frac { 1 }{ x+y } \le (\sqrt { xy } )\frac { 1 }{ 2\sqrt { xy }  } \Rightarrow \frac { \sqrt { xy }  }{ x+y } \le \frac { \sqrt { xy }  }{ 2\sqrt { xy }  } $
5) Then, $(2\sqrt { xy } )\frac { \sqrt { xy }  }{ x+y } \le (2\sqrt { xy } )\frac { \sqrt { xy }  }{ 2\sqrt { xy }  } $
6) Therefore, $\frac { 2xy }{ x+y } \le \sqrt { xy } $ 
QED

Since $x,y>0$, you can divide by any of them (or by their square roots) on both sides of the inequality. Thus,
$$\frac{2xy}{x+y}\le\sqrt{xy} \quad \Leftrightarrow \quad \frac{2\sqrt{xy}}{x+y}\le1 \quad \Leftrightarrow \quad 2\sqrt{xy}\le x+y.$$
You have already proved the rest.
