Prove this matrix inequality

Question

Let $W$ be $n$ by $n$ real symmetric positive definite matrix, let $V$ be an invertible $n$ by $n$ square matrix. Then the matrix $V^TWV$ is also symetric and positive definite. Now let $\|V\|$ denote the largest singular value of $V$. That is, the square root of the largest eigenvalue of $V^TV$.

Then, show that: \begin{equation} \min_{\|x\| = 1} x^TVWV^Tx \leq \|V\|^2 \min_{\|y\| = 1} y^TWy \end{equation}

Answer 1

Fix $x$ such that $\|x\| = 1$.

Let $y = V^\top x / \|V^\top x\|$. Note that $\|y\| = 1$ and that $\|V^\top x\| \le \|V^\top\| = \|V\|$. Then $$x^\top V W V^\top x = \|V^\top x\|^2 \cdot y^\top W y \le \|V\|^2 \cdot y^\top W y.$$