Let points $K, L, M, N$ be the vertices of the rhombus lying on sides $AE, ED, DB, BA$ respectively. Define $d(X, YZ)$ to denote the distance from point $X$ to line $YZ$. Since $D$ and $E$ are the intersection points of the angle bisectors with the opposite sides, we have $d(D, AB) = d(D, AC)$, $d(E, AB) = d(E, BC)$, and $d(D, BC) = d(E, AC) = 0$, from which we obtain
$$d(D, AC) + d(D, BC) = d(D, AB),$$
$$d(E, AC) + d(E, BC) = d(E, AB).$$
Since $L$ lies on segment $DE$, and the equation $d(X, AC) + d(X, BC) = d(X, AB)$ is linear in the variable $X$, from the above two equations we obtain
$$d(L, AC) + d(L, BC) = d(L, AB). \qquad (2)$$
Label each of the angles as shown in the figure, and let $a = KL$. Then we have $d(L, AC) = a \sin \mu$ and $d(L, BC) = a \sin \nu$. Since the parallelogram $KLMN$ lies on one side of line $AB$, it follows that
$$\begin{aligned} d(L, AB) &= d(L, AC) + d(N, BC) = d(K, AB) + d(M, AB) \\ &= a(\sin \delta + \sin \varepsilon). \end{aligned}$$
Thus, by equation (1), we obtain
$$\sin \mu + \sin \nu = \sin \delta + \sin \varepsilon. \qquad (3)$$
If one of the angles $\alpha$ and $\beta$ is not acute, then the inequality to be proved already holds. Hence we may assume $\alpha, \beta < \pi/2$.
It suffices to prove that $\psi = \angle NKL \le \max\{\alpha, \beta\}$.

We proceed by contradiction, assuming that $\psi > \max\{\alpha, \beta\}$ holds. Since $\mu + \psi = \angle CKN = \alpha + \delta$, from the assumption we obtain $\mu = (\alpha - \psi) + \delta < \delta$. Similarly, one can show that $\nu < \varepsilon$. Next, since $KN \parallel ML$, we know that $\beta = \delta + \nu$, so we have $\delta < \beta < \pi/2$. Similarly, $\varepsilon < \pi/2$. Finally, from $\mu < \delta < \pi/2$ and $\nu < \varepsilon < \pi/2$, we know that
$$\sin \mu < \sin \delta \quad \text{and} \quad \sin \nu < \sin \varepsilon.$$
These two inequalities clearly contradict (2), a contradiction. This completes the proof.