Linear matrix inequality given the bounds

Question

If $A\succeq A_{\min}\succ 0$, $B\succeq B_{\min}\succ 0$, will the following be true? $$ABA^\top \succeq AB_{\min}A^\top$$ How can I prove it?

Answer 1

We can prove by using definition of positive-definiteness. For any vector $v\in\mathbb{R}^N$ and $v\neq 0$, it's obvious that from the definition $v^\top (B - B_{\min})v \geq 0$. Since $A\succeq 0$, for any vector $x\in\mathbb{R}^N$ and $x\neq 0$, we have $A^\top x\neq 0$ because $A$ has full rank. Thus $x^\top A(B - B_{\min})A^\top x>0$. Therefore $A(B - B_{\min})A^\top\succeq 0$ and $ABA^\top\succeq AB_{\min}A^\top$.