Let $A \in \Bbb C^{n \times n}$. Show that $$\operatorname{rank}(A) \geq \frac{|\text{tr}(A)|^2}{\text{tr}(A^*A)},$$ and that the equality holds if and only if $A=\alpha H$ for some nonzero $\alpha \in \mathbb C$ and nonzero Hermitian $H$ with $\lambda(H) = \{0,1\}$.
The special case where $A$ is Hermitian was treated before:
However, this problem pertains to any complex square matrix and the equality part of this problem is non-trivial.
I have edited my answer according to @user1104082's comment, since indeed the rank could possibly be bigger than the number of nonzero eigenvalues.
Let us say that $A$ is $n$ by $n$ and that its rank is $r$.
WLOG, using the Schur decomposition, one may assume that $A$ is upper triangular.
Let $\lambda_1, \ldots, \lambda_k$ be the nonzero eigenvalues of $A$ (which may not be distinct). Note that $r \geq k$.
Consider $v = (\lambda_1, \ldots, \lambda_k)$ and $w = (1, \ldots, 1)$ (having $k$ ones). By Cauchy-Schwarz, it follows that
$$ \lvert \langle v,w \rangle \rvert ^2 \leq \lVert v \rVert^2 \lVert w \rVert^2 .$$
This translates into
$$ | \lambda_1 + \cdots + \lambda_k |^2 \leq k \left( | \lambda_1 |^2 + \cdots +| \lambda_k |^2 \right). $$
Note that the LHS is equal to $|\operatorname{tr}(A)|^2$ and the expression inside the parentheses in the RHS is less than or equal to $\operatorname{tr}(A^* A)$. Indeed, this is so because $A$ is WLOG assumed to be upper triangular (by the Schur decomposition).
So we have proved the desired inequality. And equality holds iff $v$ is proportional to $w$ and $A$ is unitarily diagonalizable (these conditions imply that $r = k$). If $\alpha$ is the factor, then equality holds iff $A$ is unitarily diagonalizable and all the non-zero eigenvalues are equal to $\alpha \neq 0$. The last condition is equivalent to $A$ being unitarily equivalent to $\alpha D$ where $D$ is the diagonal matrix having $k$ ones and $n-k$ zeros on the diagonal. Note that any matrix which is unitarily equivalent to $D$ must be hermitian. The proof is thus complete!