Since $A$ is diagonalizable in an orthonormal basis, for every vector $x$ such that $\|x\|=1$, $x^TAx=\sum a_iy_i^2$ and $ x^TA^{-1}x=\sum a_i^{-1}y_i^2$ x^TAx=\sum_ia_iy_i^2\quad\mbox{and}\quad x^TA^{-1}x=\sum_ia_i^{-1}y_i^2, $$ where the $a_i$ are the eigenvalues of $A$ and $y$ is a vector such that $\|y\|=1$. Hence $x^TAx=E(\xi)$ and $ x^TA^{-1}x=E(\xi^{-1})$ x^TAx=E(\xi)\quad\mbox{and}\quad x^TA^{-1}x=E(\xi^{-1}), $$ where $\xi$ is a random variable such that $P(\xi=a_i)=y_i^2$ for every $i$, in the simple case when the $a_i$ are distinct. (If some $a_i$ are equalIn the general case, then for every $\xi=a$ with probability the sum of the a$, $y_i^2$ such that $ a_i=a$.) P(\xi=a)=\sum_iy_i^2\ \mathbf{1}_{a_i=a}. $$ One is left with the task to prove that $E(\xi)E(\xi^{-1})\le(M+m)^2/(4Mm)$ for every random variable $\xi$ such that $m\le\xi\le M$ almost surely... This is a true inequality but I am not sure that writing it as a variance-covariance inequality is the fastest easiest way to prove it , hence Kantorovich is to write it as a variance-covariance inequality.
Recall that
To prove Kantorovich inequalitycan be proved by noting , note that, since every eigenvalue of $A$ is between $m$ and $M$, one has $A+mMA^{-1}\le(M+m)I$. $ A+MmA^{-1}\le(M+m)I. $$ Thus $a+b\le c$ with $a=x^TAx$, $ b=mM(x^TA^{-1}x)$ and a=x^TAx,\qquad b=Mm(x^TA^{-1}x),\qquad c=M+m. $c=M+m$. $ But $a+b\le c$ implies that $ab\le c^2/4$ ab\le\frac14c^2$ and this is Kantorovich inequality written as $mM(x^TAx)(x^TA^{-1}x)\le(M+m)^2/4$.$ Mm(x^TAx)(x^TA^{-1}x)\le\frac14(M+m)^2. $$
