Q:

Source/explanation for this matrix inequality

Here it is:
$$z^\top M^{-1} M^{-1}z \le \|M^{-1}\| z^\top M^{-1} z.$$
Where $\pmb M$ is positive definite symmetric, $z$ is a vector in $\mathbb{R}^p$ (not necessarily normed!) and $||\pmb M||$ is the induced operator norm, i.e., $\sup_{∥x∥≤1}∥Mx∥$

A:

HINT:
Let $A= M^{-1}$, positive definite. You want to show that
$$||Av||^2 = \langle v, A^2 v\rangle \le ||A|| \cdot \langle v, A v \rangle$$
Let $B$ the root of $A$. The inequality is equivalent to 
$$||B^2 v||^2 \le ||B^2|| \cdot ||Bv||^2$$
or 
$$||B^2 v|| \le \sqrt{||B^2||} \cdot ||Bv||$$. 
Note that for a hermitian operator $B$ we have $||B^2|| = ||B||^2$ ( in general only $\le $). 
Now it's easy. 

