Let $A \in \Bbb S^{n}$ symmetric matrices and $b \in \Bbb R^n$.
If $\|A\|_2 \le \|b\|_2$, then $|b^TAb| \le \|b\|_2^3$,
where $\|A\|_2$ is the maximum singular value of $A$ and $\|b\|_2$ is the $2$-norm of $b$.
I see that $|b^TAb| \le \|b\|_2 \|Ab\|_2$ by the Cauchy-Schwartz inequality, but from here I'm having trouble.
I also see that $\|b\|_2 \|A\|_2 \|b\|_2 \le \|b\|_2^3$, but I'm not sure how to relate $\|b\|_2 \|A\|_2 \|b\|_2$ and $|b^TAb|$.
You are working with a certain vector norm and its associated matrix norm, in the sense that $$ \|A\| = \sup_{x \ne 0}\frac{\|A x\|}{\|x\|}. $$
In this case it is easy to show that $\| A x\| \leq \|A\| \|x\|$, which allows you to say that $$ |b^T A b | \leq \|b\| \|A b\| \leq \|b| \|A\| \|b\| \leq \| b\|^3. $$