6.226. $\left\{\begin{array}{l}\sqrt{x+\frac{1}{y}}+\sqrt{y+\frac{1}{x}}=2 \sqrt{2}, \\ \left(x^{2}+1\right) y+\left(y^{2}+1\right) x=4 x y\end{array}\right.$

The system of equations is:
\[
\left\{\begin{array}{l}
\sqrt{x+\frac{1}{y}}+\sqrt{y+\frac{1}{x}}=2 \sqrt{2}, \\
\left(x^{2}+1\right) y+\left(y^{2}+1\right) x=4 x y
\end{array}\right.
\]
## Solution.

Domain of definition: $\left\{\begin{array}{l}x>0, \\ y>0 .\end{array}\right.$

## From the condition we have

$$
\left\{\begin{array} { l } 
{ \sqrt { \frac { x y + 1 } { y } } + \sqrt { \frac { x y + 1 } { x } } = 2 \sqrt { 2 } , } \\
{ x ^ { 2 } y + y + y ^ { 2 } x + x = 4 x y }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
\frac{x y+1}{y}+2 \sqrt{\frac{(x y+1)^{2}}{x y}}+\frac{x y+1}{x}=8, \\
x^{2} y+y+y^{2} x+x=4 x y
\end{array} \Leftrightarrow\right.\right.
$$

$$
\Leftrightarrow\left\{\begin{array}{l}
\frac{x^{2} y+x+x y^{2}+y}{x y}+2 \sqrt{\frac{(x y+1)^{2}}{x y}}=8, \\
x^{2} y+x+x y^{2}+y=4 x y
\end{array}\right.
$$

$$
\sqrt{\frac{(x y+1)^{2}}{x y}}=2 \Rightarrow \frac{(x y+1)^{2}}{x y}=4 \Leftrightarrow(x y-1)^{2}=0, \text { from which } x y=1 .
$$

Substituting this value of $x y$ into the first equation of the system, we get $x+x+y+y+2 \sqrt{4}=8$, from which $\left\{\begin{array}{l}x y=1, \\ x+y=2 .\end{array}\right.$ Then $\left\{\begin{array}{l}x=1, \\ y=1 .\end{array}\right.$

Answer: $(1 ; 1)$.