Frobenius Norm and Relation to Eigenvalues I've been working on this problem, and I think that I almost have the solution, but I'm not quite there.

Suppose that $A \in M_n(\mathbb C)$ has $n$ distinct eigenvalues $\lambda_1... \lambda_n$. Show that $$\sqrt{\sum _{j=1}^{n} \left | {\lambda_j} \right |^2 } \leq \left \| A \right \|_F\,.$$

I tried using the Schur decomposition of $A$ and got that $\left \| A \right \|_F = \sqrt{TT^*}$, where $A=QTQ^*$ with $Q$ unitary and $T$ triangular, but I'm not sure how to relate this back to eigenvalues and where the inequality comes from.

You are in the right way. The corresponding Schur decomposition is $A = Q U Q^*$, where $Q$ is unitary and $U$ is an upper triangular matrix, whose diagonal corresponds to the set of eigenvalues of $A$ (because $A$ and $U$ are similar). Now because Frobenius norm is invariant under unitary matrix multiplication:
$$||QA||_F = \sqrt{\text{tr}((QA)^*(QA))} = \sqrt{\text{tr}(A^*Q^* QA)} = \sqrt{\text{tr}(A^*A)} = ||A||_F$$
(the same remains for multiplication of $Q$ on the right) then we could write:
$$||A||_F = ||Q U Q^*||_F = ||U||_F \rightarrow \sqrt{\sum_{j=1}^n |\lambda_j|^2} \leq ||A||_F$$
directly proves your statement.
Note: The inequality comes from the definition of the Frobenius norm: The sum of the square of all entries in the matrix. Since $U$ contains the eigenvalues on his diagonal, the term in the left has to be less or equal to the sum over all entries, because $U$ could have non zero entries over his diagonal.
