We work in $\mathbb{R}^n$. Let $M$ be an $n\times n$ matrix with $ x^TMx \geq k\Vert x\Vert^2 $ for all $x \in \mathbb{R}^n$, where $k>0$.
I want to show that $\Vert M^{-1} \Vert \leq \frac{1}{k}$ and that the real parts of the (possibly complex) eigenvalues of $M$ are at least $k$.
Attempt so far: It is easy to see that $M$ is invertible since otherwise there would be an $x\neq 0$ with $Mx=0$ and hence $x^TMx = 0$ which would violate the condition above.
Furthermore if $\lambda$ is an eigenvalue, then $|\lambda| \geq k$ by similar reasoning. I don't see how to deal with the complex eigenvalues at all though, since this bound doesn't ignore the imaginary part.
I know that $\Vert M^{-1} \Vert = \sqrt{\rho((M^{-1})^T(M^{-1})}$ is the square root of the norm of the maximum eigenvalue of $(M^{-1})^TM^{-1}$. But since $M$ isn't necessarily symmetric I don't know how much help that will be.
Any help would be greatly appreciated.