The corners of the rhombus are named $K(K \in A E), L$ $(L \in A B), M(M \in B D)$, and $N(N \in D E)$. Let $d(X, Y Z)$ denote the distance of a point $X$ from a line $Y Z$. Since $D$ and $E$ lie on the respective angle bisectors, we have $d(D, A B)=d(D, A C)$, $d(E, A B)=d(E, B C)$, and $d(D, B C)=d(E, A C)=0$, from which $d(D, A C)+d(D, B C)=d(D, A B)$ and $d(E, A C)+d(E, B C)=d(E, A B)$ follow.
Since $N$ lies on the segment $D E$ and because in the equation $d(X, A C)+d(X, B C)=d(X, A B)$, if $X$ moves along the segment $D E$, all terms change only linearly, it follows from the two upper relations also for $N$:
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from the two upper relations also for $N$:
$d(N, A C)+d(N, B C)=d(N, A B)(1)$.
With the notations of the figure, we have $d(N, A C)=s \cdot \sin \mu$ and $d(N, B C)=s \cdot \sin v$. Since the rhombus $K L M N$ lies entirely in one of the half-planes with respect to $A B$, we obtain from its parallelogram property
$d(N, A B)=d(N, A B)+d(L, A B)=d(K, A B)+d(M, A B)=s(\sin \delta+\sin \varepsilon)$.
With (1), it follows that $\sin \mu+\sin v=\sin \delta+\sin \varepsilon$ (2).
From the assumption $\varphi>\max (\alpha, \beta)$, we would directly get $\mu=\alpha-\varphi+\delta<\delta$ and analogously $v<\varepsilon$ because $\mu+\varphi=\angle C K L=\alpha+\delta$. Since $K L \| M N$, we have $\beta=\delta+v$, so $\delta<\beta<90^{\circ}$. Similarly, it follows that $\varepsilon<90^{\circ}$. Thus, we get $\sin \mu<\sin \delta$ and $\sin v<\sin \varepsilon$, in contradiction to (2). Therefore, $\varphi \leq \max (\alpha, \beta)$.

Hint: A pure angle chase does not lead to the goal. The position of the rhombus is not uniquely determined, contrary to the opinion of some participants. Equality holds only for $\alpha=\beta$.