A recent related posting ask for conditions on equality in the Kantorovich inequality. I think is more appropriate to have a solution in this (older) posting and address the question here and the issue of equality.
The approach here is a probabilistic generalization of the result of the OP.
Theorem: Suppose $X$ is a random variable on $(\Omega,\mathscr{F},\mathbb{P})$ taking values in an interval $[a,b]$ with $0<a<b<\infty$. Then,
$$\begin{align}
1\leq \mathbb{E}[X]\mathbb{E}[X^{-1}]\leq \frac{(a+b)^2}{4ab}
\tag{0}\label{kan}\end{align}$$
Equality on the left-hand-side happens iff $X$ is constant $\mathbb{P}$-a.s; equality on the right-handside happens iff $\mathbb{P}[X=a]=\mathbb{P}[X=b]=\frac12$.
Proof:
Since $\phi(x)=\frac{1}{x}$ is convex in $(0,\infty)$, the left-side inequality follows from Jensen's inequality. Equality iff either $\phi(X)$ is linear $\mathbb{P}$-a.s. or $X$ constant a.s.
The right-side inequality follows from the Cauchy-Schwartz inequality. First notice that for any part of square integrable random variables $X,Y$
$$\left|\mathbb{E}\big[(X-\mathbb{E}(X))(Y-\mathbb{E}[Y])\big]\right| \leq\big(E\big[(X-\mathbb{E}(X))^2\big]\big)^{1/2}\big(E\big[(Y-\mathbb{E}(Y))^2\big]\big)^{1/2}$$
Equality iff either one of the random variables is a constant $\mathbb{P}$-a.s., or if there is $c>0$ such that $|X-\mathbb{E}[X]|=c|Y-\mathbb{E}[Y]|$. This is a well known result in Probability: the covariance of two random variables is at most the product of the variances of the random variables.
Taking $Y=X^{-1}$, we get
$$\mathbb{E}[X]\mathbb{E}[X^{-1}]\leq 1+\Big(E\big[\big(X-\mathbb{E}[X]\big)^2\big]\Big)^{1/2}\Big(E\big[\big(X^{-1}-\mathbb{E}[X^{-1}]\big)^2\big]\Big)^{1/2}$$
Since $0<a\leq X\leq b$, we also have that $0<\frac1b\leq\frac{1}{X}\leq \frac1a$. Recall that the mean $E[Z]$ of any square integrable random variable $Z$ satisfies $E\big[(Z-\mathbb{E}[Z])^2\big]=\inf_{a\in\mathbb{R}}E\big[(Z-a)^2\big]$. In particular,
\begin{align}
\mathbb{E}\big[\big(X-\mathbb{E}[X]\big)^2\big]&\leq \mathbb{E}\big[\big(X-\frac{a+b}{2}\big)^2\big]\leq\Big(\frac{b-a}{2}\Big)^2\\
E\big[\big(X^{-1}-\mathbb{E}[X^{-1}]\big)^2\big]&\leq \mathbb{E}\big[\big(X^{-1}-\frac{a^{-1}+b^{-1}}{2}\big)^2\big]\leq\Big(\frac{b-a}{2ab}\Big)^2
\end{align}
Putting things together, we obtain
$$
\mathbb{E}[X]\mathbb{E}[X^{-1}] \leq 1+\frac{b-a}{2}\frac{b-a}{2ab}=\frac{(a+b)^2}{4ab}
$$
The conditions for equality in the theorem follow by considering the cases where equality occur in Jensen's inequality and in Cauchy-Schwartz inequality. I leave that for anybody interested in verifying the conditions.
The inequality in the OP is a particular case of the Theorem outlined above. First notice that since $A$ is positive definite matrix, we may assume, without loss of generality, that $A$ is a diagonal matrix with positive entries $m=\lambda_1\leq\ldots\leq \lambda_n=M$. Furthermore, it is enough to consider $x$ with $\|x\|^2_2=1=\sum^n_{k=1}|x_k|^2$. Define $\Omega=\{1,\ldots,k\}$, $\mathscr{F}$ to collection of all subsets of $\Omega$, and $\mathbb{P}[\{k\}]=|x_k|^2$, and $X(k)=\lambda_k$.
A survey with other interesting extensions can be found in
Bühler, W. Two proofs of the Kantorovich Inequality and generalizations, Revista Colombiana de Matemáticas, Vol. 21 (1987), pp.147-154.
