Let $K, L, M$, and $N$ be the vertices of the rhombus lying on the sides $A E, E D, D B$, and $B A$, respectively. Denote by $d(X, Y Z)$ the distance from a point $X$ to a line $Y Z$. Since $D$ and $E$ are the feet of the bisectors, we have $d(D, A B)=d(D, A C), d(E, A B)=d(E, B C)$, and $d(D, B C)=d(E, A C)=0$, which implies
$$d(D, A C)+d(D, B C)=d(D, A B) \quad \text{ and } \quad d(E, A C)+d(E, B C)=d(E, A B) .$$
Since $L$ lies on the segment $D E$ and the relation $d(X, A C)+d(X, B C)=d(X, A B)$ is linear in $X$ inside the triangle, these two relations imply
$$\begin{equation*} d(L, A C)+d(L, B C)=d(L, A B) \tag{1} \end{equation*}$$
Denote the angles as in the figure below, and denote $a=K L$. Then we have $d(L, A C)=a \sin \mu$ and $d(L, B C)=a \sin \nu$. Since $K L M N$ is a parallelogram lying on one side of $A B$, we get
$$d(L, A B)=d(L, A B)+d(N, A B)=d(K, A B)+d(M, A B)=a(\sin \delta+\sin \varepsilon)$$
Thus the condition (1) reads
$$\begin{equation*} \sin \mu+\sin \nu=\sin \delta+\sin \varepsilon . \tag{2} \end{equation*}$$

If one of the angles $\alpha$ and $\beta$ is non-acute, then the desired inequality is trivial. So we assume that $\alpha, \beta<\pi / 2$. It suffices to show then that $\psi=\angle N K L \leqslant \max \{\alpha, \beta\}$.
Assume, to the contrary, that $\psi>\max \{\alpha, \beta\}$. Since $\mu+\psi=\angle C K N=\alpha+\delta$, by our assumption we obtain $\mu=(\alpha-\psi)+\delta<\delta$. Similarly, $\nu<\varepsilon$. Next, since $K N \| M L$, we have $\beta=\delta+\nu$, so $\delta<\beta<\pi / 2$. Similarly, $\varepsilon<\pi / 2$. Finally, by $\mu<\delta<\pi / 2$ and $\nu<\varepsilon<\pi / 2$, we obtain
$$\sin \mu<\sin \delta \quad \text{ and } \quad \sin \nu<\sin \varepsilon$$
This contradicts (2).