Determine all positive real numbers $x$ and $y$ satisfying the equation

$$
x+y+\frac{1}{x}+\frac{1}{y}+4=2 \cdot(\sqrt{2 x+1}+\sqrt{2 y+1}) .
$$
Answer: $x=y=1+\sqrt{2}$.

Note that

$$
x+\frac{1}{x}+2-2 \sqrt{2 x+1}=\frac{x^{2}+2 x+1-2 x \sqrt{2 x+1}}{x}=\frac{1}{x}(x-\sqrt{2 x+1})^{2} .
$$

Hence the original equation can be rewritten as

$$
\frac{1}{x}(x-\sqrt{2 x+1})^{2}+\frac{1}{y}(y-\sqrt{2 y+1})^{2}=0 .
$$

For $x, y>0$ this gives $x-\sqrt{2 x+1}=0$ and $y-\sqrt{2 y+1}$. It follows that the only solution is $x=y=1+\sqrt{2}$.