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.