In a paper, the author removed the matrix $P$ and use the maximum eigenvalue multiplied by identity matrix , so is the following true?
$$x^T P x \le x^T \bar\lambda(P) I x$$ where $x\in\mathbb R^n$, $P \succ 0$ is positive definite, $\bar\lambda(P)$ is the largest eigenvalue of $P$ and $I$ is the identity matrix
If yes, can you give me a proof or a reference for that?
Yes. This follows from the characterization of positive semidefinite matrices as those for which all eigenvalues are nonnegative. If $\lambda$ is the maximum eigenvalue of a real symmetric (or more generally hermitian) matrix $P$, then the eigenvalues of $\lambda I - P$ are the differences $\lambda - \mu$ for eigenvalues $\mu$ of $P$, and therefore are all nonnegative. So $\lambda I - P$ is positive semidefinite, which says $x^T (\lambda I - P) x \ge 0$.