Notation

Let $I_n$ be an $n$th order unit matrix.

Problem

We want to show that$$\operatorname{tr}(I_n-\Lambda) + \log \det \Lambda \leq 0,$$where $n\times n$ matrix $\Lambda$ is a positive definite matrix.

What I know

For any $\lambda >0$, the inequality$$\log \lambda \leq \lambda - 1$$is valid.Furthermore,$$\log\det A = \operatorname{tr}\log A$$is also known to hold.In this case, the LHS of the inequality we want to show can be transformed as follows:$$\operatorname{tr}(I_n-\Lambda) + \log \det \Lambda = \operatorname{tr}(I_n-\Lambda) + \operatorname{tr}\log \Lambda = \operatorname{tr}(I_n-\Lambda + \log\Lambda).$$

I am almost ready to show it, but what should I do from here?