5
$\begingroup$

for $x,y>0$ prove that

$\frac{2xy}{x+y}\leq \sqrt{xy}\leq \frac{x+y}{2}$

I have tried to develop $(x+y)^2=$ and to get to an expression that must be bigger than those above

Thanks!

$\endgroup$
  • $\begingroup$ I hope you tried something, please tell us what you tried? $\endgroup$ – AD. Oct 28 '13 at 20:47
  • $\begingroup$ I saw that both sides are built from (x+y) so I try do develop an equality with (x+y)^2 when I can prove that one side is equal to the other, and therefore it will be bigger from what that is asked for $\endgroup$ – gbox Oct 28 '13 at 22:04
1
$\begingroup$

For example

$$\frac{2xy}{x+y}\le\sqrt{xy}\iff 4x^2y^2\le xy(x^2+2xy+y^2)\iff$$

$$ \iff xy(x^2+2xy+y^2-4xy)\ge 0\iff xy(x-y)^2\ge 0$$

And since the last rightmost inequality is obvious we're done.

$\endgroup$
4
$\begingroup$

$\sqrt{xy} \le \dfrac{x + y}{2}$ is just the AM–GM inequality. $\dfrac{2xy}{x + y} \le \sqrt{xy}$ is also the AM-GM inequality applied to $1/x$, $1/y$.

$\endgroup$
0
$\begingroup$

For the 1st one , $(x+y)/2 \ge \sqrt{xy}$

$x+y\ge 2*\sqrt{xy}$

$(\sqrt{x})^2+(\sqrt{y})^2-2*\sqrt{xy}\ge0$

$(\sqrt{x}-\sqrt{y})^2\ge0$ ---->>> DONE

For the 2nd one :

$x+y\ge 2xy/\sqrt{xy}$

$x+y\ge 2*\sqrt{xy}$ --> this is the one just proved above by divide both side by 2.

$\endgroup$
0
$\begingroup$

Start by noticing that since $(x-y)^2 \geq 0$ then $x^2 - 2xy +y^2 \geq 0$ which implies that $x^2+y^2 \geq 2xy$. Now we attack these inequalities one at a time.

To show the left inequality, we can show $2xy \leq (x+y)\sqrt{xy}$, and since everything is positive we can square both sides and further reduce the problem to showing $4x^2 y^2 \leq(x^2+2xy+y^2)(xy)$. Then, since $x^2+y^2 \geq 2xy$ we have $(x^2+2xy+y^2)(xy) \geq 4xy(xy) = 4x^2y^2$, which is exactly what you wanted to show.

To show the right inequality we will square both sides again and reduce the problem to showing $xy \leq \frac{1}{4}(x^2 + 2xy +y^2)$. To do so we again use that $x^2+y^2 \geq 2xy$. Therefore, $\frac{1}{4}(x^2+2xy+y^2) \geq \frac{1}{4}(4xy) = xy$ which is again the inequality you wanted to show.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.