Prove $\text{rank}(A) \geq \frac{(\text{tr}(A))^2}{\text{tr}(A^2)}$ when $A$ is Hermitian If $A \in \mathbb{C}^{n \times n}$ is a non-zero Hermitian matrix, I need to show that 
$$\text{rank}(A) \geq \frac{(\text{tr}(A))^2}{\text{tr}(A^2)}$$ and reason out when the equality is attained?  
Can anyone help me in showing this result?
 A: Your matrix is diagonalizable to a diagonal matriz with real entries, when you diagonalize it, $\mathrm{rank}A$, $\mathrm{tr} A$ and $\mathrm{tr} A^2$ do not change. This means that you can suppose that $A$ is real and diagonal to begin with. 
So suppose $A=\operatorname{diag}(\lambda_1,\dots,\lambda_n)$ with the $\lambda_i\in\mathbb R$. Then $\operatorname{rank}A$ is the number of non-zero $\lambda_i$s, $\operatorname{tr}A=\sum_i\lambda_i$ and  $\operatorname{tr}A^2=\sum_i\lambda_i^2$. It follows that we are reduced to showing that

if $\lambda_1$, $\dots$, $\lambda_n$ are real numbers, then $\big(\sum_{i=1}^n\lambda_i\big)^2\leq n\sum_{i=1}^n\lambda_i^2$.

One way to do prove this is to fix $r\geq0$, compute the maximum $M$ of the function $f(\lambda_1,\dots,\lambda_n)=(\lambda_1+\cdots+\lambda_n)^2$ in the sphere $\lambda_1^2+\cdots+\lambda_n^2=r$, and finally check that $M$ is equal to $nr$.
This can be done easily using the method of Lagrange multipliers---it will, moreover, tell you where the maximum is achieved, and show when your inequality becomes an equality.
