Example 5 Given that $a$ and $b$ are positive numbers, and $a+b=2$. Find the minimum value of $u=\sqrt{a^{2}+1}+\sqrt{b^{2}+4}$.
(2003, Beijing Middle School Mathematics Competition (Grade 8))
Explanation: If we consider this problem from an algebraic perspective, it would be quite challenging. By observation, it is not difficult to find that the conditions in the problem have a clear geometric background.

Let's consider $\sqrt{a^{2}+1}$ and $\sqrt{b^{2}+4}$ as the hypotenuses of right triangles with legs $a, 1$ and $b, 2$ respectively, and construct the geometric figure shown in Figure 1, where $A C \perp A B, B D \perp A B$, and $A C=1, B D=A B=2$. Then,
$$
P C=\sqrt{a^{2}+1}, P D=\sqrt{b^{2}+4} .
$$

To find the minimum value of $u$, the problem can be transformed into: finding a point $P$ on the line segment $A B$ such that $P C+P D$ is minimized.

Construct the symmetric point $C^{\prime}$ of point $C$ with respect to the line $A B$, and connect $C^{\prime} D$ intersecting $A B$ at point $P$. By the property of the shortest distance between two points in plane geometry, point $P$ is the desired point.

By symmetry, it is clear that $P C=P C^{\prime}$. Draw $C^{\prime} D^{\prime} \perp D B$, intersecting the extension of $D B$ at point $D^{\prime}$. Then,
$$
\begin{array}{l}
P C+P D=P C^{\prime}+P D=C^{\prime} D \\
=\sqrt{C^{\prime} D^{\prime 2}+D^{\prime} D^{2}}=\sqrt{A B^{2}+(A C+B D)^{2}} \\
=\sqrt{13} .
\end{array}
$$

Therefore, the minimum value of $u$ is $\sqrt{13}$.