Prove that $\kappa_F(A) \ge \sqrt{n}$

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I reached at a point after solving it

$$\kappa_F(A) = \sqrt{\mbox{tr}(A^H A) \cdot \mbox{tr}(A^{-1}(A^{-1})^H}$$

Now I am stuck. How to proceed? Or, alternatively, a new approach is also appreciated.

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Hint. Every complex square matrix $X$ is unitarily triangulable and Frobenius norm is unitarily invariant. It follows that $\|X\|_F^2\ge\sum_i|\lambda_i(X)|^2$.