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If $x,y$ are positive real numbers, then find the maximum value of :

$\frac{xy}{x+y}$

Is there any specific inequality to use here?

Don't solve it, just direct me

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    $\begingroup$ If you don't have constrains on $x$ and $y$, $(x,y)\to\infty$ $\endgroup$ – Riccardo.Alestra Nov 21 '13 at 15:31
  • $\begingroup$ We need to maximize $xy$ & minimize $x+y$ , but $x+y\ge 2\sqrt{xy}$ $\endgroup$ – lab bhattacharjee Nov 21 '13 at 15:31
  • $\begingroup$ Fool around a bit with numbers, no calculator. It will hit you. Hard. $\endgroup$ – André Nicolas Nov 21 '13 at 15:42
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What happens if you put $x=y$?

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  • $\begingroup$ how do you establish that $x=y$ gives the maximum value of the given expression? $\endgroup$ – lab bhattacharjee Nov 21 '13 at 15:33
  • $\begingroup$ @lab: It doesn't. That's not the point. $\endgroup$ – TonyK Nov 21 '13 at 15:40
  • $\begingroup$ @lab: I've only observed that if you put $x=y$, your expression becomes $x^2/(2x)=x$, that is not upperly bounded. What does it means? $\endgroup$ – gangrene Nov 21 '13 at 15:42
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Maximizing $\frac{xy}{x+y}$ is equivalent to minimizing $\frac{1}{x} + \frac{1}{y}$. Perhaps, that helps?

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Consider $\left\langle\frac{\partial}{\partial x}\left(\frac{xy}{x+y}\right),\frac{\partial}{\partial y}\left(\frac{xy}{x+y}\right)\right\rangle$

We can express $\frac{xy}{x+y}$ as $y\left(1-\frac{y}{x+y}\right)$, which could make the partial derivative easier to calculate.

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