Inner product inequality for symmetric matrix I could use some guidance regarding this question.

Let $A$ be an $n\times n$ symmetric matrix with non-negative elements. Prove that for any nonzero $x \in \mathbb{R}^n$ with non-negative elements the following inequality holds:
$$
\Bigg(\frac{\langle x, Ax\rangle}{\langle x, x\rangle}\Bigg)^m \leq \frac{\langle x, A^mx\rangle}{\langle x, x\rangle}, \quad m \in \mathbb{Z}^+
$$
where $\langle \cdot, \cdot \rangle$ denotes the inner (dot) product. Hint: Use induction.

All I have so far is this after the inductive step:
$$
\Bigg(\frac{\langle x, Ax\rangle}{\langle x, x\rangle}\Bigg)^k \Bigg(\frac{\langle x, Ax\rangle}{\langle x, x\rangle}\Bigg) \leq \Bigg(\frac{\langle x, A^kx\rangle}{\langle x, x\rangle}\Bigg) \Bigg(\frac{\langle x, Ax\rangle}{\langle x, x\rangle}\Bigg)
$$ for some $k \in \mathbb{Z}^+$, trying to show the inequality for $k+1$. I messed around with the symmetric matrix fact and transposes on the RHS, but I did not get anywhere.
I asked around at the department, but I did not get much help. A few questions right away:

*

*What fundamental thing am I missing here? There must be some piece of information or some overall concept that I am not seeing. These look like Rayleigh quotients (although I do not know much about them). Are they key to solving this problem?


*What does non-negative entries give you? Is it simply to ensure that you don't divide by zero, or is there something else going on?


*I know that each $x$ in the statement is essentially a unit vector. Does this have something to do with the problem? When you apply $A$ though, I do not think that $x$ remains a unit vector in general.
UPDATE(1): I'm pretty sure this has to do with the spectra of real symmetric matrices. I just looked up Rayleigh quotients, and it seems like there are some theorems more generally that involve forms like those in this question.
UPDATE(2): I was given a suggestion (not a hint, so I do not know yet if it is fruitful) to try to prove the case for $m=2$ and then prove the inductive step of $m$ implies $m + 2$. I have not yet had the time to try this, but it seems promising.
 A: For $m=2$ use Cauchy-Schwarz inequality with $x$ and $Ax$
$$\left\langle x, Ax\right\rangle^2 \le \left\langle Ax, Ax\right\rangle \left\langle x,x\right\rangle = \left\langle x, A^2x\right\rangle\left\langle x, x\right\rangle.$$

For general case the answer is not complete
Now assume the statement is true for $m$, since $Ax$ has also non-negative entries if $Ax=0$ or $\left\langle x ,Ax\right\rangle = 0$, the inequality is trivial, otherwise you will have:
\begin{align}
\frac{\left\langle x, A^{m+2}x\right\rangle}{\langle x, x\rangle} &= \frac{\left\langle Ax, A^{m}Ax\right\rangle}{\langle Ax, Ax\rangle} \frac{\langle Ax, Ax\rangle}{\langle x, x\rangle}\\
&\ge \left(\frac{\left\langle Ax, A Ax\right\rangle}{\left\langle Ax, Ax\right\rangle}\right)^m\frac{\left\langle x, A^2x\right\rangle}{\left\langle x, x\right\rangle}\\
&= \left(\frac{\left\langle x, A^3x\right\rangle}{\left\langle x, A^2x\right\rangle}\right)^m\frac{\left\langle x, A^2x\right\rangle}{\left\langle x, x\right\rangle}\\
&\ge \left(\frac{\left\langle x, A^3x\right\rangle}{\left\langle x, A^2x\right\rangle}\right)^m\left(\frac{\left\langle x, Ax\right\rangle}{\left\langle x, x\right\rangle}\right)^2\\
&= \left(\frac{\left\langle x, A^3x\right\rangle\left\langle x, x\right\rangle}{\left\langle x, A^2x\right\rangle\left\langle x, Ax\right\rangle}\right)^m\left(\frac{\left\langle x, Ax\right\rangle}{\left\langle x, x\right\rangle}\right)^{m+2}\\
\end{align}
Moreover let $P = \frac1{\left\langle x, x\right\rangle} xx^T$,
\begin{align}
\left\langle x,A^2x\right\rangle\left\langle x, Ax\right\rangle &= \left\langle x, x\right\rangle x^TA^2PAx\\
&= \frac12\left\langle x, x\right\rangle\left(x^T\left(A^2PA + APA^2\right) x\right)\\
&= \frac12\left\langle x, x\right\rangle\left(x^TA\left(AP + PA\right)A x\right)
\end{align}
To complete the proof I think one should prove that $$\left\langle x, A^3x\right\rangle \left\langle x, x\right\rangle \ge \left\langle x, A^2x\right\rangle \left\langle x, Ax\right\rangle.$$
