If symmetric matrix $A\geq0$, $P>0$, does $APA\leq \lambda_{max}^2(A) P$ always hold?

Question

If symmetric matrix $A\geq0$, $P>0$, can $APA\leq \lambda_{max}^2(A) P$ always hold?

Notation:

$\lambda_{max}(A)$ means matrix $A$'s largest eigenvalue.

$A\geq0$ means matrix A is a positive semi-definite matrix.

$P>0$ means matrix P is a positive definite matrix.

Answer 1

No, it is not always, the case. For example $$A=\left(\begin{array}{cc}1&1\\1&1\end{array}\right), P=\left(\begin{array}{cc}100&0\\0&1\end{array}\right)$$

Then $$APA = \left(\begin{array}{cc}101&101\\101&101\end{array}\right)\not\le 4P$$