Let $X, V\in\mathbb{R}^{n\times r}$ such that $X^\top V$ is symmetric. The central quantity I care about is
\begin{equation}
\|XV^\top\|\_{F}^2+\|X^\top V\|\_{F}^2 +[\text{Tr}(X^\top V)]^2.
\end{equation}
An easy lower bound for this quantity is given by $2\sigma\_{r}(X)^2\|V\|\_{F}^2$, where $\sigma\_{r}(X)$ is the smallest singular value of $X$. 


I'm wondering whether there exists some absolute constant $c>0$ such that the following holds
\begin{equation}
\|XV^\top\|\_{F}^2+\|X^\top V\|\_{F}^2 +[\text{Tr}(X^\top V)]^2\geq c\|X\|\_{F}^2\|V\|\_{F}^2.
\end{equation}



Nontrivial lower bound on the sum of matrix norms
No. With $n = r = 2$, set $$X = \bigg(\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array} \bigg) \, , \quad V = \bigg( \begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array} \bigg) \, .$$
In particular, $X^T V = V^T X = 0$, the zero matrix.


If you restrict to invertible square matrices, the statement is still false. Set
$$X = \bigg(\begin{array}{cc} 1 & 0 \\ 0 & \epsilon \end{array} \bigg) \, , \quad V = \bigg( \begin{array}{cc} \epsilon & 0 \\ 0 & 1 \end{array} \bigg) \, .$$
Then $X^T V = V^T X = \epsilon \operatorname{Id}$ and $\operatorname{Tr}(X^T V) = 2 \epsilon$. So, your LHS is
$$
2 \| \epsilon \operatorname{Id}\|\_F^2 + (\operatorname{Tr}(\epsilon \operatorname{Id}))^2 = 8 \epsilon^2 \, ,
$$
which can be made arbitrarily small, while your RHS is
$$
c \| X \|\_F^2 \| V \|\_F^2 = c (1 + \epsilon^2)^2 \geq c 
$$


The moral of the story is that the matrix norm is submultiplicative (not the Frobenius norm as above, but bear with me) in the sense that $\| X V \|$ can be arbitrarily smaller than $\| X \| \| V \|$. In the end, your constant $c$ depends either on how $X, V$ map each others singular value spaces to each other, or on some lower bound on the lowest singular value for either $X$ or $V$.


