Example 1 If $x, y$ are positive numbers, then the minimum value of $\left(x+\frac{1}{y}\right)^{2}+\left(y+\frac{1}{2 x}\right)^{2}$ is
A. $4 \sqrt{2}$
B. $2+2 \sqrt{2}$
C. $3+2 \sqrt{2}$
D. $6 \frac{1}{4}$
Solve $\left(x+\frac{1}{y}\right)^{2}+\left(y+\frac{1}{2 x}\right)^{2}=\left(x^{2}+\frac{1}{4 x^{2}}\right)+\left(\frac{2 x}{y}+\frac{y}{x}\right)+\left(y^{2}+\frac{1}{y^{2}}\right)$.
Given $x>0, y>0$, we have
$$
\begin{array}{c}
x^{2}+\frac{1}{4 x^{2}} \geqslant 2 \sqrt{x^{2} \cdot \frac{1}{4 x^{2}}}=1 ; \\
\frac{2 x}{y}+\frac{y}{x} \geqslant 2 \sqrt{\frac{2 x}{y} \cdot \frac{y}{x}}=2 \sqrt{2} ; \\
y^{2}+\frac{1}{y^{2}} \geqslant 2 \sqrt{y^{2} \cdot \frac{1}{y^{2}}}=2 ;
\end{array}
$$
(1) + (2) + (3), we get $\left(x+\frac{1}{y}\right)^{2}+\left(y+\frac{1}{2 x}\right)^{2} \geqslant 1+2 \sqrt{2}+2=3+2 \sqrt{2}$.

Equality holds if and only if (1), (2), and (3) all hold as equalities. That is, when $x=\frac{\sqrt{2}}{2}, y=1$, the inequality (4) holds as an equality. Therefore, the correct answer is $\mathrm{C}$.