Let $$A$$ be any real square matrix (not necessarily symmetric).  Prove that: $$(x'A x)^2 \leq (x'A A'x)(x'x)$$

The key point in proving this inequality is to recognize that $$x'A A'x$$ can be expressed as vector norm of $$A'x$$.

Proof:

If $$x=0$$, then the inequality is trival.

Suppose $$x \neq 0$$.

$$\frac{x'A x}{x'x} = \frac{(A'x)'x}{\| x \|^2} = (A'\frac{x}{\| x \|})'\frac{x}{\| x \|}$$

Because $$\frac{x}{\| x \|}$$ is a unit vector, $$A'\frac{x}{\| x \|}$$ can be considered as scale and rotation of $$\frac{x}{\| x \|}$$ by $$A'$$.  Thus, the resulting vector norm of $$A'\frac{x}{\| x \|}$$ is $$\alpha$$ for some $$\alpha > 0$$.  And $$(A'\frac{x}{\| x \|})'\frac{x}{\| x \|}=\alpha \, cos(\beta)$$ for some $$-\pi \leq \beta \leq \pi$$, which is the angle between before and after premulitplying $$A'$$.

Now:

$$( \frac{x'A x}{x'x} )^2$$

$$= ( (A'\frac{x}{\| x \|})'\frac{x}{\| x \|} )^2$$

$$=\alpha^2 \, cos(\beta)^2$$

$$\leq \alpha^2$$

$$= (A'\frac{x}{\| x \|})'A'\frac{x}{\| x \|}$$

$$= \frac{(A'x)'A'x}{\| x \|^2}$$

$$= \frac{x'A A'x}{x'x}$$

Finally, multiplying both sides by $$(x'x)^2$$ completes the proof.