Introduce the following assumptions and definitions,
- $\Omega\subset \mathbb{R}^3$ a bounded Lipshitz domain,
- $(L^2(\Omega))^3$ the space of vector square integrable functions,
- $M$ a $3\times 3$ real symmetric positive definite matrix (so $M$ is invertible),
- $A: (L^2(\Omega))^3\to (L^2(\Omega))^3$ a bounded self adjoint operator which satisfies: $$ AM\neq MA $$ and $$ 0\leq (u,Au)\leq \| u\|^2 ,\;\forall u\in (L^2(\Omega))^3, $$ where $(\cdot,\cdot)$ is the $L^2$ scalar product and $\|\cdot\|$ is the associated norm. Then, I want to check if we have or not the following property: $$ 0\leq (u,(M^{-1}AM)u)\leq \| u\|^2 ,\;\forall u\in (L^2(\Omega))^3. $$ I know that the operator $M^{-1}AM$ defined on $(L^2(\Omega))^3$ is spectrally equivalent to $A$ but I don't know how to use it to check the above inequalities.
This already fails in the scalar case. Consider $$ A=\begin{bmatrix} 0&0\\0&1\end{bmatrix},\qquad\qquad M=\begin{bmatrix} 1&1\\1&2\end{bmatrix}. $$ Both are selfadjoint and invertible, and $\|A\|=1$, so $(u,Au)\leq \|u\|^2$. We have $$ M^{-1}AM=\begin{bmatrix} 2&-1\\-1&1\end{bmatrix}\begin{bmatrix} 0&0\\0&1\end{bmatrix} \begin{bmatrix} 1&1\\1&2\end{bmatrix} =\begin{bmatrix} -1&-2\\1&2\end{bmatrix}. $$ Then $$ \bigg(\begin{bmatrix} 0\\1\end{bmatrix} ,M^{-1}AM\begin{bmatrix} 0\\1\end{bmatrix} \bigg)=\bigg(\begin{bmatrix} 0\\1\end{bmatrix},\begin{bmatrix} -2\\2\end{bmatrix}\bigg)=2>1=\bigg\|\begin{bmatrix} 0\\1\end{bmatrix}\bigg\|^2. $$
Edit: controlling $M$ is not enough.
Let $$ A=\begin{bmatrix} 0&0\\0&1\end{bmatrix},\qquad\qquad M=\begin{bmatrix} 1/2&1/4\\1/4&1/2\end{bmatrix}. $$ Then $\|M\|=3/4$ and $$ M^{-1}AM=\frac13\begin{bmatrix} -1&-2\\ 2&4\end{bmatrix}, $$ and $$ \bigg(\begin{bmatrix} 0\\1\end{bmatrix} ,M^{-1}AM\begin{bmatrix} 0\\1\end{bmatrix} \bigg)=\bigg(\begin{bmatrix} 0\\1\end{bmatrix},\begin{bmatrix} -2/3\\4/3\end{bmatrix}\bigg)=\frac43>1=\bigg\|\begin{bmatrix} 0\\1\end{bmatrix}\bigg\|^2. $$