Suppose $\mathcal K_1, \mathcal K_2 \subset M_{m \times n}(\mathbb R)$ are two connected compact sets both containing $0$ and for every nonzero $M \in \mathcal K_1$ and $N \in \mathcal K_2$, we have $M^{\top}N + N^{\top}M$ to be positive definite. I want to estimate a lower bound for the trace of $M^{\top}N + N^{\top}M$ in terms of trace of $M^{\top}M$.
If we take out any arbitrary open ball $\mathcal B_{\varepsilon}(0)$ about $0$ in $\mathcal{K_1, K_2}$, as trace is a continuous function, $\operatorname{Trace}(M^{\top}N + N^{\top}M)$ achieves minimum $c_1$ over $(\mathcal K_1 \setminus \mathcal B_{\varepsilon}(0)) \times (\mathcal K_2 \setminus \mathcal B_{\varepsilon}(0))$ and $\operatorname{Trace}(M^{\top}M)$ achieves maximum $c_2$ over $\mathcal K_1 \setminus \mathcal B_{\varepsilon}(0)$. We may take $a= \frac{c_1}{c_2}$. Is it possible to make a global estimate for $a$?
If $0\in\mathcal{K_2}$, then, with $N=0$, it holds $0=Tr(M^\top N + N^\top M)\geq a Tr(M^\top M)$. So either $a \leq 0$ or $M = 0$ for all $M\in \mathcal{K_1}$.
Therefore, to have $\mathcal{K_1}\neq \{0\}$ and $a>0$ you have to exclude an open neighborhood of the origin in $\mathcal{K_2}$. And in this case, your proof of $a=c_1/c_2$ works well for existence.