Solve for positive reals Solve for positive reals $x,y$
$(x+y)(1+\frac{1}{xy})+4=2(\sqrt{2x+1}+\sqrt{2y+1})$
I started by accumulating the terms of $x$ and then used AM-GM inequality but unsuccessfully....
 A: 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.
A: 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.
