I am looking for an elementary proof of the following:
Let $x$ be a $k\times 1$ column vector and $A$ be a square symmetric matrix of order $k$, with all elements of $x$ and $A$ non-negative. Then for any positive integer $n$,
$$(x^\top A^n x)(x^\top x)^{n-1}\geqslant(x^\top Ax)^n$$
with equality if and only if $x$ is an eigenvector of $A$.
I prefer not to do this by induction. So it felt like I have to use the Cauchy-Schwarz inequality. But then I was looking at the result $(x^\top Ax)(y^\top Ay)\geqslant (x^\top Ay)^2$ for a positive definite matrix $A$, which is seemingly nowhere near the desired result.
I then looked at the equality case to see if I can do this.
If $\lambda$ is an eigenvalue of $A$ corresponding to the eigenvector $x$, then we have
$Ax=\lambda x\tag{1}$
Premultiplying $(1)$ by $x^\top$, we get $x^\top Ax=\lambda x^\top x$.
From $(1)$ we again have $A^nx=\lambda^n x$, so that $x^\top A^nx=\lambda^nx^\top x$.
So, the l.h.s of the desired inequality becomes $$(x^\top A^n x)(x^\top x)^{n-1}=\lambda^n(x^\top x)^n$$ while the r.h.s is simply $(x^\top Ax)^n=\lambda^n(x^\top x)^n$.
But I need a hint on how to prove the strict inequality in general. I wonder if any results from extrema of quadratic forms would come in handy.
Apparently this inequality occurs somewhere in genetical theory (Mulholland and Smith, 1959). There is a proof of this inequality in the cited paper but I haven't gone through it.