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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. 
Since $A$ is diagonalizable in an orthonormal basis, for every vector $x$ such that $\|x\|=1$, $x^TAx=\sum a_iy_i^2$ and $x^TA^{-1}x=\sum a_i^{-1}y_i^2$ where the $a_i$ are the eigenvalues of $A$ and $y$ is a vector such that $\|y\|=1$. Hence $x^TAx=E(\xi)$ and $x^TA^{-1}x=E(\xi^{-1})$ 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 equal, then $\xi=a$ with probability the sum of the $y_i^2$ such that $a_i=a$.) 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 way to prove it, hence Kantorovich inequality.
Recall that Kantorovich inequality can be proved by noting that, since every eigenvalue of $A$ is between $m$ and $M$, one has $A+mMA^{-1}\le(M+m)I$. Thus $a+b\le c$ with $a=x^TAx$, $b=mM(x^TA^{-1}x)$ and $c=M+m$. But $a+b\le c$ implies that $ab\le c^2/4$ and this is Kantorovich inequality written as $mM(x^TAx)(x^TA^{-1}x)\le(M+m)^2/4$.