here $A$ is a Hermitian square matrix.
A few thoughts on the problem:
We know the matrix $A$ is unitarily diagonalizable, so $A = PDP^{-1}$ for some unitary matrix $P$.
Since this is the only assumption given in the problem, the above factorization is probably crucial, but I do not know how to utilize it.
So, I am instead trying some brute force calculations on the RHS of the inequality, with not much success.
For instance, expanding the numerator in the RHS and then regrouping terms, I see that
$$[\operatorname{tr}(A)]^2 = \sum_i^n a_{ii}^2 + 2\sum_{i \ne j}a_{ii}a_{jj}$$
I am not sure how to expand the denominator, $tr(A^2)$, without more information about $A$.
Thanks in advance,
Hint: If $A$ is similar to the diagonal matrix $D$, then $A$ and $D$ will have the same rank and trace. Moreover, $A^2$ will be similar to $D^2$.
Alternatively, note that since $A$ is Hermitian, $$ \operatorname{trace}(A^2) = \sum_{i=1}^n \sum_{j=1}^n |a_{ij}|^2 $$
Suppose without loss of generality that $A$ is diagonal of the form $$ A = \pmatrix{\lambda_1\\ &\ddots \\ && \lambda_r \\ &&&0 \\ &&&& \ddots\\ &&&&& 0} $$ where the eigenvalues $\lambda_1,\dots,\lambda_r$ are real and non-zero.
Note that rank$(A) = r$. Now, the tricky bit: we apply the Cauchy-Schwartz inequality. We have $$ \operatorname{trace}(A)^2 = \left[\sum_{j=1}^r\lambda_j\right]^2 = \\ |(\lambda_1,\dots,\lambda_r)\cdot(1,\dots,1)|^2 \leq\\ \|(\lambda_1,\dots,\lambda_r)\|^2 \|(1,\dots,1)\|^2 = \\ \left[\sum_{j=1}^r\lambda_j^2\right] \cdot r = \operatorname{rank}(A) \operatorname{trace}(A^2) $$ The result follows.