Let $\mathbf{A}$ be a full column-rank matrix with unit $\ell_2$-norm columns, and let $\mathbf{v}_{1}, \cdots, \mathbf{v}_{r}$ be the vectors such that $\{ \mathbf{A} \mathbf{v}_{1}, \cdots, \mathbf{A} \mathbf{v}_{r} \}$ is orthonormal. Also, let $c_{1}, \cdots, c_{r}$ be real numbers satisfying $c_{1}^{2}+ \cdots + c_{r}^{2} = 1$.
In this case, does the following inequality hold true? \begin{align} \underset{i=1}{\overset{r}{\sum}} \left\| \mathbf{A}^{T} \mathbf{A} \mathbf{v}_{i} \right\|_{2}^{2} &\ge \frac{r}{\left\| \underset{i=1}{\overset{r}{\sum}} c_{i} \mathbf{v}_{i} \right\|_{2}^{2}} \end{align}
Clearly, the above inequality holds for $r=1$, because
$$\| \mathbf{A}^{T} \mathbf{A} \mathbf{v}_{1} \|_{2}^{2} \cdot \| \mathbf{v}_{1} \|_{2}^{2} \ge \langle \mathbf{A}^{T} \mathbf{A} \mathbf{v}_{1}, \mathbf{v}_{1} \rangle^{2} = \| \mathbf{A} \mathbf{v}_{1} \|_{2}^{4} = 1.$$
No. Take $A = \begin{bmatrix}1 & 0\\0 & \epsilon\end{bmatrix}$ and $V = A^{-1}$.
Then $$\sum_{i=1}^2 \|A^TA\mathbf{v}_i\|^2 = 1 + \epsilon^2$$ but we can set $c_0=1, c_1=0$ to get $$\frac{r}{\left\|\sum_{i=1}^2 c_i\mathbf{v}_i\right\|^2}=2$$ violating the inequality for $\epsilon$ sufficiently small.
What I believe is true is $$\sum_{i=1}^r \left\|A^TA\mathbf{v}_i\right\|^2 \geq \frac{1}{r\left\|\sum_{i=1}^r c_i \mathbf{v}_i\right\|^2}.$$
This is easy to see when $r$ is equal to the rank of $A$, since in that case the left-hand side is equal to $\operatorname{tr}(AA^T)$ and the right-hand side is bounded above by $\lambda_{\mathrm{max}}(AA^T)/r.$ When $r$ is less than the rank of $A$, I believe the same argument will work, this time restricting $AA^T$ to the subspace spanned by the $Av_i$.