Example 5 Given $a, \beta \in\left(0, \frac{\pi}{2}\right)$, and $\cos \alpha+\cos \beta-$ $\cos (\alpha+\beta)=\frac{3}{2}$. Find the values of $\alpha, \beta$.
Solving the original condition can be transformed into
$$\sin \alpha \sin \beta+(1-\cos \alpha) \cos \beta=\frac{3}{2}-\cos \alpha$$

From $\cos ^{2} \beta+\sin ^{2} \beta=1$, we can get
$$\frac{\sin ^{2} \alpha \sin ^{2} \beta}{\sin ^{2} \alpha}+\frac{(1-\cos \alpha)^{2} \cos ^{2} \beta}{(1-\cos \alpha)^{2}}=1,$$

Therefore, by Theorem 1, we have
$$\sin ^{2} \alpha+(1-\cos \alpha)^{2} \geqslant\left(\frac{3}{2}-\cos \alpha\right)^{2},$$

Simplifying, we get $\left(\cos \alpha-\frac{1}{2}\right)^{2} \leqslant 0$, thus $\cos \alpha=\frac{1}{2}, \alpha=$ $\frac{\pi}{3}$, and by the symmetry of $a, \beta$, we know $\beta=\frac{\pi}{3}$.