The question is an extention to the answered question prove spectral equivalence bounds for fractional power of matrices.

Let $A, D \in \mathbb{R}^{n \times n}$ be two symmetric,positive definite and tri-diagonal matrices for that we know that they are spectrally equivalent, thus ist holds $$ c^- x^\top D x \le x^\top A x \le c^+ x^\top D x $$ for any $x \in \mathbb{R}^n$, where $c^+, c^- > 0.$ The matrices $A$ and $D$ can be diagonalized, that is $$ A = V\Lambda_A V^\top, \quad D = W\Lambda_D W^\top $$ where $V$ and $W$ contain the eigenvectors of $A$ and $D$, and $\Lambda_A$ and $\Lambda_D$ are diagonal matrices containing the respective eigenvalues. Based on the Reileigh quotient, it should follow that $$ cond(D^{-1}A) \le \frac{c^+}{c^-},$$ thus $c^+$ and $c^-$ upper and lower bounds for the range of the eigenvalues of $D^{-1}A.$

In the question I linked above, I got the answer that due to Loewner's theorem, for $0 < \alpha \le 1,$ $$ (c^-)^\alpha x^\top D^\alpha x \le x^\top A^\alpha x \le (c^+)^\alpha x^\top D^\alpha x $$ does hold. Here, $A^\alpha := V\Lambda_A^\alpha V^\top,$ and $D^\alpha := W\Lambda_D^\alpha W^\top,$ where $\Lambda_A^\alpha, \Lambda_D^\alpha$ can be computed by taking the power $\alpha$ of each diagonal entry.

**Now my question is:**

Is it possible to deduce the spectral bound estimates for the inverse of the matrices, that is for $A^{-1}$ and $D^{-1}$ as well as $A^{-\alpha}$ and $D^{-\alpha}$ in the same manner ? I expect something like, e.g., $$\frac{1}{c^+} x^\top D^{-1} x \le x^\top A^{-1} x \le \frac{1}{c^-} x^\top D^{-1} x $$ and $$\frac{1}{(c^+)^\alpha} x^\top D^{-\alpha} x \le x^\top A^{-\alpha} x \le \frac{1}{(c^-)^\alpha} x^\top D^{-\alpha} x $$ that holds for any $x \in \mathbb{R}^n$.

And second: Is it possible to deduce the condition numbers for $(D^{-1} A)$ and $(D^{-\alpha} A^\alpha)$ from that?