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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

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    $\begingroup$ Hint: Take reciprocals in $\sqrt{xy}\leq\frac{x+y}{2}$. $\endgroup$ – Guest Oct 23 '15 at 10:57
  • $\begingroup$ It's good. I sugest to improve (shorten) at step 4, multiplying both sides by $2xy$ instead of $\sqrt{xy}$, so step $5$ won't be necessary. $\endgroup$ – Bernard Oct 23 '15 at 12:21
  • $\begingroup$ But then it would be $\frac { 2xy }{ x+y } \le \frac { xy }{ \sqrt { xy } } $ How is that the same? $\endgroup$ – Cherry_Developer Oct 23 '15 at 12:33
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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.

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The left-hand side inequality (HGM inequality) is simply the right-hand side (AGM inequality), applied to $\dfrac1x$ and $\dfrac1y$, reversed.

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  • $\begingroup$ I will add what I did to finish the proof to my original post. Please look it over. $\endgroup$ – Cherry_Developer Oct 23 '15 at 11:37

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