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

$$A$$ 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.

• Shouldn't the first inequality be $\langle Ax, x \rangle^4 \leq ||Ax||^4 ||x||^4$? Mar 30 '20 at 4:10

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