What are the possible solutions of $x+y+ {1\over x}+{1\over y}+4=2 (\sqrt {2x+1}+\sqrt {2y+1})$?

It's $$\sum_{cyc}\left(x+\frac{1}{x}+2-2\sqrt{2x+1}\right)=0$$ or $$\sum_{cyc}\frac{x^2-2x\sqrt{2x+1}+2x+1}{x}=0$$ or $$\sum_{cyc}\frac{(x-\sqrt{2x+1})^2}{x}=0,$$ which for $xy<0$ gives infinitely many solutions.

But, for $xy>0$ we obtain: $$x=\sqrt{2x+1}$$ and $$y=\sqrt{2y+1},$$ which gives $$x=y=1+\sqrt2.$$

This solution works only for positive $x,y$. However they can not be both negative since then LHS is at most $0$. So $$x+y+ {1\over x}+{1\over y}+4 = x+y+{2x+1\over x}+{2y+1\over y} $$

By Am-Gm we have $$ x+{2x+1\over x}\geq 2\sqrt{x{2x+1\over x}} = 2\sqrt{2x+1}$$ and the same for $y$, so we have

$$x+y+ {1\over x}+{1\over y}+4 \geq 2\sqrt{{2x+1}}+2\sqrt{{2y+1}}$$

Since we have equality is achieved when $x={2x+1\over x}$ (and the same for $y$) we have $x=y=1+\sqrt{2}$