In my homework, I've made it to the point where I obtained this value
$$\frac{(v^Tv)^2}{v^TQvv^TQ^{-1}v}$$
where $v \in \mathbb{R}^n$ and $Q\in \mathbb{R}^{n\times n}$ and $Q$ is symmetric positive definite.
I know I have to use Kantorovich's inequality:
For $v_1, \ldots, v_n >0 \ : \ v_1 \leq v_2 \leq \ldots \leq v_n$ and $w_1, ... w_n \geq 0 \ : \ w_1 + w_2 + \cdots w_n=1$, then, $$(\sum w_i v_i)(\sum w_i v_i^{-1}) \leq \frac{(v_1+v_n)^2}{4v_1v_n}$$
I am confused with how to proceed though. I know one of either $v$'s or $w$'s is the elements of the sum $v^Tv$. My main problem is finding what is $w_i$ and $v_i$
Let $Q=U^TD U$ be the symmetric eigenvalue decomposition with sorted eigenvalues. Let $r=Uv$,
\begin{align} \frac{(v^Tv)^2}{v^TQvv^TQ^{-1}v} &= \frac{(v^Tv)^2}{v^TU^TD Uvv^TU^TD^{-1} Uv} \\ &=\frac{(r^Tr)^2}{r^TD rr^TD^{-1} r} \\ &= \frac{(r^Tr)^2}{\left(\sum_{i=1}^nd_ir_i^2\right)\left(\sum_{i=1}^nd_i^{-1}r_i^2\right)} \end{align}
Hopefully you can take it from here.