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

\begin{align*}
\frac { 2xy }{ x+y } \le \sqrt { xy } \le \frac { x+y }{ 2 }\\
\text{assuming $x,y>0, \space$ we square all terms }\\
\frac{4x^2y^2}{(x+y)^2 } \le xy  \le \frac{(x+y)^2}{4}
\\
\end{align*}
\begin{align*}
\text{we eliminate denominators }
\\
(4x^2y^2(4)  &=16x^2y^2  \\
\le xy(4)(x+y)^2  &=4 x^3 y + 8 x^2 y^2 + 4 x y^3\\
\le 
(x+y)^2(x+y)^2&=x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4
\\
 \\
\text{subtract }\space 4x^2y^2+4x^3y+4xy^3\space& \text{ from all sitermses}
 \\
\space -4 x y (x^2 - 3 x y + y^2) 
&\le (2xy)^2 \le  (x^2+y^2)^2
\end{align*}
