Find the minimum value of n

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.

• I guess $n$ is a natural number here – George Dewhirst May 9 at 11:56
• Yes the minimum value of n should be 4 – Suman Chandra May 9 at 11:57
• 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? – Andreas Caranti May 9 at 12:00
• 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. – George Dewhirst May 9 at 12:01
• @GeorgeDewhirst, got it, thx. I should have noticed the tags. – Andreas Caranti May 9 at 12:03

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.