6. Let $x, y$ be positive real numbers, and satisfy $x+y=1$. Find the minimum value of $\left(\frac{x+1}{x}\right)^{2}+\left(\frac{y+1}{y}\right)^{2}$.
6. By the arithmetic-geometric mean inequality, we have
$$
1=x+y \geqslant 2 \sqrt{x y}, \text{ hence } \frac{1}{x y} \geq 4.
$$
$$
\begin{array}{l}
\left(\frac{x+1}{x}\right)^{2}+\left(\frac{y+1}{y}\right)^{2} \\
\geqslant \frac{1}{2}\left(\frac{x+1}{x}+\frac{y+1}{y}\right)^{2} \\
=\frac{1}{2}\left(1+\frac{1}{x y}\right)^{2} \geqslant \frac{25}{2},
\end{array}
$$

Equality holds if and only if $x=y=\frac{1}{2}$.