I am looking at the negative log-likelihood of a multivariate Gaussian $X \sim \mathcal{N}(\mu, \Sigma)$: $$const + (n/2)\log|\Sigma| + (1/2)Tr(\Sigma^{-1}S),$$ where $S = (1/n)\sum(X_i-\mu)(X_i-\mu)'$ is the sample covariance matrix.
And I wonder can we bound the $Tr(\Sigma^{-1}S)$ term with an expression involving only $Tr(\Sigma^{-1})$. I read somewhere[1] in SE that if $A, B$ are two real symmetric $n \times n$ PSD matrices, then: $$Tr(AB) \le \lambda_{max}(B)Tr(A),$$ where $\lambda_{max}(B)$ is the maximal eigenvalue of $B$.
If true, since $\Sigma^{-1}, S$ are real symmetric $p \times p$ PSD matrices, this would mean: $$Tr(\Sigma^{-1}S) \le \lambda_{max}(S)Tr(\Sigma^{-1})$$
With the help of simulations, I can never find a case where this isn't true. Two main questions:
(1) How do I prove this? (hopefully a process I would understand...)
(2) Is there a better/interesting upper bound involving $Tr(\Sigma^{-1})$ which I'm missing?
[1] No idea where it was, otherwise I would probably start from there...