Lower bound on smallest eigenvalue of (symmetric positive-definite) matrix

If $M$ is a positive-definite symmetric matrix, is it possible to get a positive lower bound on the smallest eigenvalue of $M$ in terms of a matrix norm of $M$ or elements of $M$? Eg. I want $$\lambda_{\text{min}} \geq f(\lVert M \rVert)$$ or something like that.

$M$ is a Gram matrix if that helps. Thanks.

There is one lower bound on minimum eigenvalue of symmetric p.d. matrix given in [Applied Math. Sc., vol. 4, no. 64] which is based on Frobenius norm (F) and Euclidean norm (E)

$$ \lambda_{min} \gt \sqrt{\frac{||A||_F^2-n||A||_E^2}{n(1-||A||_E^2/|det(A)|^{2/n})}} $$

if it helps.

[reference]: K. H. Schindler, "A New Lower Bound for the Minimal Singular Value for Real Non-Singular Matrices by a Matrix Norm and Determinant", Journal of Applied Mathematical Sciences, Vol. 4, No. 64, 2010.

A quick comment: If you have diagonal dominance, then Gerhsgorin's circle theorem for eigenvalues will get you at least something. So for each row, subtract the diagonal term from the sum of the absolute values of the off-diagonal terms, and take the minimum over the rows. That is a bound on the eigenvalue that will be positive (again, if you have diagonal dominance, which may not hold for all Gram matrices).