The following inequality intuitively holds in my opinion, however I am facing hard time proving it
${\lambda _n}\left( {{X^T}AX} \right) \leqslant \left\| X \right\|_F^2{\lambda _n}\left( A \right)$
Keep in mind that $\lambda_n$ is the smallest eigenvalue of a $n$-by-$n$ square matrix. The matrix $A$ is a symmetric, positive definite matrix
All I know from matrix linear algebra is that
${\lambda _n}\left( {{X^T}AX} \right) \leqslant {\lambda _1}\left( {{X^T}AX} \right) = \rho \left( {{X^T}AX} \right) \leqslant {\left\| {{X^T}AX} \right\|_F} \leqslant \left\| X \right\|_F^2{\left\| A \right\|_F}$
where $\rho$ is the spectral radius.