Given rectangular matrix $A \in \mathbb{R}^{n\times m}$ and positive definite matrix $B \in \mathbb{R}^{n \times n}$, prove that $$-\mbox{tr}(A^TBA) \leq -\sigma\|A\|^2_F$$ where constant $\sigma > 0$ may depend on $A$ and $B$.
I have developed a rudimentary proof for the same, however I want to know if my reasoning is correct or if someone can improve upon my proof or provide a better proof. My proof is as follows: Since $A^TBA$ is a positive definite matrix, $tr(A^TBA) \geq ||A^TBA||_2$, therefore: \begin{equation} \begin{split} tr(A^TBA) &\geq ||A^T||_2||B||_2||A||_2 \\ -tr(A^TBA) &\leq -\rho(B)||A||^2_F/r \end{split} \end{equation} In my reasoning above, I have utilized the fact that $||A||_2 \leq ||A||_F \leq \sqrt{r}||A||_2$ where $r$ is rank of $A$ matrix. My main contention is if $tr(A^TBA) \geq ||A^T||_2||B||_2||A||_2$ holds? Also $tr$ represents trace and $\rho(.)$ represents spectral radius and thus according to my 'not-so-rigorous' proof $\sigma=\rho(B)/r$. Can any expert provide a better/concrete proof for $-tr(A^TBA) \leq -\sigma||A||^2$. Lastly, thanks for your time and efforts!
Let $B=Q\Lambda Q^T$ be an orthogonal diagonalisation. By the tracial property $\operatorname{tr}(XY)=\operatorname{tr}(YX)$, we get \begin{aligned} \operatorname{tr}(A^TBA) &=\operatorname{tr}(A^TQ\Lambda Q^TA)\\ &=\operatorname{tr}(\Lambda Q^TAA^TQ)\\ &\ge\lambda_\min(B)\operatorname{tr}(Q^TAA^TQ)\\ &=\lambda_\min(B)\operatorname{tr}(AA^T)\\ &=\lambda_\min(B)\|A\|_F^2. \end{aligned} So, you may pick $\sigma$ as the minimum eigenvalue of $B$.
The inequality $\operatorname{tr}(A^TBA)\ge\|A^T\|_2\|B\|_2\|A\|_2$ is wrong. E.g. for every $0<t<1$, we have $$ \operatorname{tr}(A^TBA):=\pmatrix{0&1}\pmatrix{1\\ &t}\pmatrix{0\\ 1}=t<1=\|A^T\|_2\|B\|_2\|A\|_2. $$