Let $A \in M_{n}(\mathbb{R})$,$\ \alpha_1,\alpha_2,\cdots,\alpha_n\ $are the column vector of $A.\ $Proof$:\ $rank($A$)$\geq \sum_\limits{i=1}^{n}\dfrac{a_{ii}^{2}}{\alpha_{i}^{T}\alpha_{i}}$.
I show my opinion. I assumed that the matrices have rank r,$\alpha_{1},\cdots,\alpha_{r}$ are extremely uncorrelated groups and then tried to split the summation into two parts $[1,r]$ and $[r+1,n]$. For the latter part I have $\alpha_j$ linearized by $\alpha_{1},\cdots,\alpha_{r}$, but it doesn't seem to work.
Let $A$ be rank $k$, so we can write $A = QR$ for some $Q\in\mathbb R^{n\times k}$ with orthonormal columns. Now let $r_i\in\mathbb R^k$ denote the columns of $R$. Then, we have that $\|r_i\|_2 = \|Re_i\|_2 = \|QRe_i\|_2 = \|Ae_i\|_2 = \|\alpha_i\|_2$, where we have used that matrices with orthonormal columns preserve the $\ell_2$ norm. Then, \begin{align*} \sum_{i=1}^n \frac{\alpha_{ii}^2}{\|\alpha_i\|_2^2} &= \sum_{i=1}^n \frac{\langle e_i^\top Q, Re_i\rangle^2}{\|\alpha_i\|_2^2} \\ &\leq \sum_{i=1}^n \frac{\|e_i^\top Q\|_2^2 \|r_i\|_2^2}{\|\alpha_i\|_2^2} && \text{Cauchy-Schwarz} \\ &= \sum_{i=1}^n \|e_i^\top Q\|_2^2 && \text{$\|r_i\|_2 = \|\alpha_i\|_2$} \\ &= \|Q\|_F^2 = \sum_{j=1}^k \|Qe_j\|_2^2 = k \end{align*}