Here is a longish, but simple proof (some ideas taken from Prof. Bhatia's book on positive definite matrices, but adapted to match the setting of the paper you cite).

To simplify notation, assume without loss of generality that all matrices are real and symmetric. Then, the cited paper defines the *covariance* for an arbitrary unit vector $x$ as

$$C(A,B) = x^TBAx - (x^TAx)(x^TBx),$$

using which it further defines $V(A) := C(A,A)$.

The standard covariance-variance inequality is

$$(*)\hskip 10pt |C(A,B)| \le \sqrt{V(A)}\sqrt{V(B)}.$$

Using the above inequality, here is one way in which we can prove the Kantorivich inequality.

First, define $\delta=(x^TAx)(x^TA^{-1}x)$; then, observe that $C(A,A^{-1}) = 1 - \delta$.

Inequality (*) says that

$$|1-\delta| \le \sqrt{V(A)}\sqrt{V(A^{-1})}.$$

So let us first bound $V(A)$ and $V(A^{-1})$. Assume therefore $mI \preceq A \preceq MI$ (here $\preceq$ denotes the L\"owner ordering).

Notice that $(MI-A)(A-mI) \succeq 0$, or in other words
\begin{eqnarray*}
A^2 + mMI &\preceq& (m+M)A\\\\
x^TA^2x-(x^TAx)^2 + mM &\le& x^TAx[(m+M) - x^TAx]\\\\
x^TA^2x-(x^TAx)^2 + mM &\le& \frac{1}{4}(m+M)^2\\\\
x^TA^2x-(x^TAx)^2 &\le& \frac{1}{4}(M-m)^2.
\end{eqnarray*}
Similarly, we obtain $V(A^{-1}) \le \frac{1}{4}(1/m-1/M)^2$.

Finally, since $\delta \ge 1$, we have that $|1-\delta| = \delta - 1$.
Putting the pieces together we obtain from (*)
$$\delta - 1 \le \frac{1}{4}(M-m)(1/m-1/M),$$
which you can simplify to finish the proof.