If $A$ is skew-symmetric, then $x^T A^2 x \le 0$

Question

Let $A$ be a skew-symmetric matrix. Show that for any vector $x$

$$x^T A^2 x \le 0$$

Since $A$ is skew-symmetric then $A^2$ is symmetric. Then, I'm blocked. I tried to consider $a=x^T A^2 x$, but without success.

Answer 1

Hint: If $A$ is skew-symmetric, then $x^TA = -x^TA^T$. Therefore $$x^TA^2x = -x^TA^TAx = -(x^TA^T)(Ax) = \ldots\ ?$$