Example 7 Given a cube $A B C D-A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ with edge length 1, take a point $P$ on edge $A C$, and construct a plane through points $P, A^{\prime}, B^{\prime}$ that forms a dihedral angle $\alpha$ with the base. Construct another plane through points $P, B^{\prime}, C^{\prime}$ that forms a dihedral angle $\beta$ with the base. Find the minimum value of $\alpha+\beta$.
Solution: Draw $P P^{\prime} \perp A^{\prime} C^{\prime}$ at $P^{\prime}$, it is easy to know that $P P^{\prime} \perp$ the base $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$.

Draw $P^{\prime} N \perp A^{\prime} B^{\prime}$, then $P N \perp A^{\prime} B^{\prime}$. Similarly, draw $P^{\prime} M \perp A^{\prime} B^{\prime}$ to get $P M \perp B^{\prime} C^{\prime}$. Rotate $\triangle P P^{\prime} N$ around $P P^{\prime}$ to make it fall into the plane $P A^{\prime} P^{\prime}$.

Rotate $\triangle P P^{\prime} M$ around $P P^{\prime}$ to the plane $P A^{\prime} P^{\prime}$, thus obtaining a new triangle $P N^{\prime} M^{\prime}$ with $\alpha, \beta$ as the two base angles.

Let $A P=a$, then $N^{\prime} P^{\prime}=N P^{\prime}=$ $\frac{a}{\sqrt{2}}, P^{\prime} M^{\prime}=P^{\prime} M=\frac{\sqrt{2}-a}{\sqrt{2}}$, so, $N^{\prime} M^{\prime}=N^{\prime} P^{\prime}+P^{\prime} M^{\prime}=1$.
From $\left\{\begin{array}{l}S=\frac{1}{2} a b \sin \alpha \\ \cos \alpha=\frac{a^{2}+b^{2}-c^{2}}{2 a b}\end{array}\right.$ we know that when
the base length and height of $\triangle P M^{\prime} N^{\prime}$ are fixed, and the other two sides are equal, the vertex angle is the largest. Therefore, only when $P$ is at the midpoint of $A C$, $\angle N^{\prime} P M^{\prime}$ is the largest, $\alpha+\beta=\pi-\angle N^{\prime} P M^{\prime}$ is the smallest. At this time, $N^{\prime} P^{\prime}=$ $\frac{1}{2}, \operatorname{tg} \alpha=2$.

Since $P$ is at the midpoint of $A C$, $\alpha=\beta$, then $\operatorname{tg}(\alpha+\beta)=\operatorname{tg} 2 \alpha=-\frac{4}{3}$, hence $(\alpha+$ $\beta)_{\min }=\pi-\operatorname{arctg} \frac{4}{3}$.