Let function $f : \mathbb R^{m \times n} \to \mathbb R_0^+$ be defined as follows

$$f (\mathrm X) := \| \,\mathrm X \mathrm A^\top \|_\text{F}^2 + \| \,\mathrm X^\top \mathrm A \,\|_\text{F}^2 + \left( \langle \mathrm A, \mathrm X \rangle \right)^2$$

where $\mathrm A \in \mathbb R^{m \times n}$ is given. Note that

$$\| \,\mathrm X \mathrm A^\top \|_\text{F}^2 = \| \,\mathrm A \mathrm X^\top \|_\text{F}^2 \geq \lambda_n ( \mathrm A^\top \mathrm A ) \, \| \mathrm X \|_\text{F}^2$$

$$\| \mathrm X^\top \mathrm A \|_\text{F}^2 = \| \mathrm A^\top \mathrm X \|_\text{F}^2 \geq \lambda_m ( \,\mathrm A \mathrm A^\top ) \, \| \mathrm X \|_\text{F}^2$$

and that $\left( \langle \mathrm A, \mathrm X \rangle \right)^2 \geq 0$. Hence,

$$f (\mathrm X) \geq \left( \lambda_n ( \mathrm A^\top \mathrm A ) + \lambda_m ( \,\mathrm A \mathrm A^\top ) \right) \| \mathrm X \|_\text{F}^2$$

Suppose that $\rm A$ is **tall** (i.e., $m > n$) and has **full column rank** (i.e., $\mbox{rank} (\mathrm A) = n$). In this case,

$$\lambda_n ( \mathrm A^\top \mathrm A ) = \sigma_n^2 (\mathrm A) = \left( \frac{\| \mathrm A \|_2}{\kappa (\mathrm A)} \right)^2$$

where $\kappa (\mathrm A)$ is the (finite) [condition number][1] of $\rm A$, and $\lambda_m ( \,\mathrm A \mathrm A^\top ) = 0$. Thus,

$$f (\mathrm X) \geq \left( \frac{1}{\kappa (\mathrm A)} \right)^2 \| \mathrm A \|_2^2 \, \| \mathrm X \|_\text{F}^2$$

which is not exactly what the question's author wanted, but is arguably (somewhat) close.


  [1]: https://en.wikipedia.org/wiki/Condition_number