How can I prove the inequality ?
$$ \|\mathbf{X}\|_\sigma \leq \sqrt{rank(X)} \|\mathbf{X}\|_F $$
where $\|\mathbf{X}\|_\sigma$ is the trace norm (sum of singular values of $\mathbf{X}$) and $\|\mathbf{X}\|_F$ is the Frobenius norm of $\mathbf{X}$.
The only thing I can see is that $\|\mathbf{X}\|_F \leq \|\mathbf{X}\|_\sigma$
Both norms are unitarily invariant. So, by singular value decomposition, you may assume that $X$ is a (possibly rectangular) diagonal matrix and the inequality in question reduces to Cauchy-Schwarz inequality about the vector containing the singular values of $X$ and the all-one vector.