$\langle Ax,x \rangle^4\leq \langle A^4x,x\rangle$

Question

$A$ is an adjacency matrix for a graph so $A$ is symmetric. $x$ is a unit vector.

Why is $\langle Ax,x\rangle^4\leq \langle A^4x,x\rangle$?

So far I have that $\langle Ax,x\rangle^4\leq ||Ax||^2||x||^2= ||Ax||^2$ By the Cauchy Schwartz inequality.

Answer 1

Hint: $\langle A^4x, x \rangle = \langle A^2x, (A^T)^2x \rangle$.

Answer 2

This is true for any Hermitian matrix $A$. By a change of orthonormal basis, we may assume that $A=\operatorname{diag}(a_1,a_2,\ldots,a_n)$ is a real diagonal matrix. Let $c_i=|x_i|^2$. Then $c_i\ge0,\,\sum_ic_i=1$ and the inequality in question is equivalent to $\left(\sum_i c_ia_i\right)^4\le\sum_ic_ia_i^4$, which is true because $f(t)=t^4$ is a convex function.