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Find the minimum $natural$ value of $n$ for given expression $$(x+y) ^2=nxy$$ I know that intuitively that the value will be minimum at x=y, but i want the mathematical logic behind it. Please don't send by doing it by $$A.M.>=G.M.$$ because I already know that method. Thanks.

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    $\begingroup$ I guess $n$ is a natural number here $\endgroup$ – George Dewhirst May 9 at 11:56
  • $\begingroup$ Yes the minimum value of n should be 4 $\endgroup$ – Suman Chandra May 9 at 11:57
  • $\begingroup$ I don't quite understand. Sure, if you use the inequality between the arithmetic and geometric mean you obtain that the minimum is $4$. What is wrong with this argument? $\endgroup$ – Andreas Caranti May 9 at 12:00
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    $\begingroup$ It is a perfectly reasonably argument I think Suman just wants to use calculus. You can find $4$ as a local minimum by taking both partial derivatives of the function $(x+y)^2/xy$ and setting equal to zero. $\endgroup$ – George Dewhirst May 9 at 12:01
  • $\begingroup$ @GeorgeDewhirst, got it, thx. I should have noticed the tags. $\endgroup$ – Andreas Caranti May 9 at 12:03
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Note that $\frac{d}{dx}(x+y)^2/xy = [(xy)(2x+2y)-y(x+y)^2]/(xy)^2$ this is zero whenever $x=y$.

This is symmetric so it is the same story with the $y$ derivative. The value of the function at $x=y$ is $4$.

We can see this is a local minimum if we plot the graph. It is also clear that for the $(x,y)$ where this function is strictly positive, this is the minimum value the function takes.

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  • $\begingroup$ Thanks a lot for your help $\endgroup$ – Suman Chandra May 9 at 12:26

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