$A_{2}^{T} + A_{2} < 0$ for $A_{2} =(A_{1}A_{0}^{-1})^{\alpha}A_{0}$?

Question

Given are two matrices $A_0, A_1$, whose symmetric part is negative definite:

$A_{0}^{T} + A_{0} < 0$,

$A_{1}^{T} + A_{1} < 0$

How could one proof that: $A_{2}^{T} + A_{2} < 0$ for

$A_{2} = (A_{1}A_{0}^{-1})^{\alpha}A_{0}$ and $\alpha \in [0, 1]$?