If $|(ax + by)(ay + bx)| \le x^2 + y^2$ holds for all real $x$ and $y$, then prove $a^2 + b^2 \le 2$

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How do I go about proving the following statement?

Let $a, b \in \mathbb{R}$. Let the inequality $$|(ax + by)(ay + bx)| \le x^2 + y^2$$ hold $\forall x, y \in \mathbb{R}$. Show that $$a^2 + b^2 \le 2$$

I have been going at this problem for an hour now and I can't seem to find a reasonable approach, I always have a "leftover" term that I can't get rid of.