Solve for positive reals $x,y$


I started by accumulating the terms of $x$ and then used AM-GM inequality but unsuccessfully....

  • $\begingroup$ Are you interested in the answer or the steps of the solution? $\endgroup$ – Paul Sundheim Aug 15 '14 at 18:28
  • $\begingroup$ Steps of course... $\endgroup$ – Satvik Mashkaria Aug 15 '14 at 18:29
  • $\begingroup$ You cannot really use AM-GM inequality, since you wish to solve an equality (though in some cases it would work). That would yield a solution only in some cases. $\endgroup$ – Theon Alexander Aug 15 '14 at 18:37

First: $$\begin{align} (x+y)\left(1+\frac{1}{xy}\right)+4 &= x+y+\frac{1}y+\frac{1}x +4\\ &= \left(x+2+\frac{1}{x}\right)+\left(y+2+\frac{1}{y}\right) \\ &=\frac{1}{x} \left(x+1\right)^2 + \frac{1}{y}\left(y+1\right)^2 \end{align}$$

So, define $$f(x)=\frac{1}{x}\left(x+1\right)^2-2\sqrt{2x+1}$$

Then you want to find two positive values $x,y$ so that $f(x)=-f(y)$. Can $f(u)$ ever be negative or zero?

Letting $g(x)=xf(x) = (x+1)^2-2x\sqrt{2x+1}$, you need to solve $g(x)\leq 0$.

So $(x+1)^4 \leq 4x^2(2x+1)$ or $$x^4-4x^3+2x^2+4x+1\leq 0$$

$x^4-4x^3+2x^2+4x+1$ factors (via Wolfram Alpha) as $(x^2-2x-1)^2$. So there are no negative values for $g$ and only one positive zero at $x=1+\sqrt 2$, and therefore a solution of $x=y=1+\sqrt 2$ to your original equation, and this is the only positive real solution.


Let's see.

Symmetric polynomials in $x,y$ are always polynomials in $x+y, xy$. The same goes with rational functions.

The problem here is that we have radicals, right? So things get more complicated.

A possibility would be the following.

Define $u, v$ as follows: $2x+1=u^2, 2y+1=v^2$, and then substitute above. Then it would be nice for you to write the equation in terms of $u+v, uv$, since it is the case that, when we isolate $.... =0$, it will be symmetric on $u,v$.

Hope this helps.


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