Does convergence of a trace imply that of an eigenvalue?

Question

I came up with the following conjecture:

Let $A_n$ be an $n \times n$ positive definite matrix, and suppose that $0 < \lambda_1^{(n)} \leq \cdots \leq \lambda_n^{(n)}$ are the eigenvalues of $A_n.$ Then $\frac{1}{n}\mathrm{tr}(A_n) \to 1$ and $\frac{1}{n}\mathrm{tr}(A_n^{-1}) \to 1$ hold if and only if $\lambda_1^{(n)} \to 1$ and $\lambda_n^{(n)} \to 1.$

The "if" part is obvious because all eigenvalues of $A_n$ get close to one, but I cannot find a good way to show the "only if" part. I would appreciate it if you could give me a hint. You can assume some additional conditions on the order of the convergence if necessary.

Answer 1

Let $ A_n$ be the diagonal matrix with diagonal entries $\frac12, 1,\ldots, 1$. Then $$\frac1n\operatorname{tr} A_n=1-\frac1{2n}\to 1$$ and $$\frac1n\operatorname{tr} A_n^{-1}=1+\frac1n\to 1$$ but of course $\lambda_1^{(n)}\not\to 1$.