On Page 490 of B & V's Convex Optimization book, by the Lipschitz condition (9.31), we have, for $t\ge 0$,
$$\tag{1} \left\|\nabla^{2} f\left(x+t \Delta x_{\mathrm{nt}}\right)-\nabla^{2} f(x)\right\|_{2} \leq t L\left\|\Delta x_{\mathrm{nt}}\right\|_{2} $$ and therefore $$\tag{2} \left|\Delta x_{\mathrm{nt}}^{T}\left(\nabla^{2} f\left(x+t \Delta x_{\mathrm{nt}}\right)-\nabla^{2} f(x)\right) \Delta x_{\mathrm{nt}}\right| \leq t L\left\|\Delta x_{\mathrm{nt}}\right\|_{2}^{3} $$ My question is how to get (2) from (1). It seems that, after mutiplying $\left\|\Delta x_{\mathrm{nt}}\right\|_{2}^{2}$ on both sides of (1), applying Cauchy-Schwartz inequality on the lefthand side is a good thought, but the basic Cauchy-Schwartz inequality is only applicable for dot product between two vectors. What can we do now?
You can use the property that $\|Ax\|_2\leq\|A\|_2\|x\|_2$. In particular, \begin{align} \|x^\top A x\| = \|x^\top (Ax)\| \leq \|x\| \|Ax\|\leq \|x\|\|A\|\|x\| \leq tL \|x\|^3 \end{align} using the given bound on $A = \nabla^2 f(\bar{x} + tx) - \nabla^2 f(\bar{x})$.