For any Hermitian matrix $\bf A$ and invertible matrix $\bf B$, how do I derive the inequality below $${\rm tr}(({\bf B}^H{\bf A}{\bf B}) \circ({\bf B}^H{\bf A}{\bf B}))\geq \lambda_{\min}^4({\bf B}){\rm tr}({\bf A}^2),$$ where $\circ$ denotes the Hadmard product.
I have changed the question.
This inequality is false. Take $\mathbf{B} = \mathbf{I}$ and $\mathbf{A} = \begin{bmatrix} 1 & 1 \\ 1 & 1\end{bmatrix}$. Then
$$ \begin{align} \mathrm{LHS} &= \operatorname{tr} \left( \begin{bmatrix} 1 & 1 \\ 1 & 1\end{bmatrix} \right) = 2, \\ \mathrm{RHS} &= \lambda_{\rm min}^4(\mathbf{I})\operatorname{tr}\left( \begin{bmatrix} 2 & 2 \\ 2 & 2\end{bmatrix} \right) = 4. \end{align} $$