I was reading Boyd's convex optimization where I encountered a version of Cauchy-Schwarz inequality that I hadn't seen before. It is used in the following context (Page 460, after 9.11 if you want to check, but it's not really necessary for the rest of this question):
\begin{equation} f(x^*) \geq f(x) + \nabla f(x)^T(x^*-x) + \frac{m}{2} \left \Vert x^*-x \right\Vert_2^2 \\ \geq f(x) - \left\Vert \nabla f(x) \right\Vert_2 \left\Vert x^* - x \right\Vert_2+ \frac{m}{2} \left \Vert x^*-x \right\Vert_2^2 \end{equation}
So I was wondering about this inequality: $a^Tb > - \left\Vert a \right\Vert_2 \left\Vert b \right\Vert_2 $ and wanted to find out if this always holds or just in this case where we can be sure that the inner product is not larger than 0. I came up with the following: Assume $a^Tb < 0$: Then with standard CS: $\left\vert a^Tb\right\vert \leq \left\Vert a \right\Vert_2 \left\Vert b \right\Vert_2$ we get: $$ a^T b = -\left\vert a^Tb\right\vert \geq -\left\Vert a \right\Vert_2 \left\Vert b \right\Vert_2.$$
If $a^Tb \geq 0$: In this case we know from the non-negativity of the norm that: $$ 0 \leq \left\Vert a \right\Vert_2 \left\Vert b \right\Vert_2 \Leftrightarrow 0 \geq -\left\Vert a \right\Vert_2 \left\Vert b \right\Vert_2, $$ so we can conclude that $a^Tb \geq 0 \geq -\left\Vert a \right\Vert_2 \left\Vert b \right\Vert_2$.
After writing this down here it feels kind of trivial, however, I just wanted to make sure that I got this one right as I normally only see people using CS for lower bounds.
Thank's for your help!