$A \in \mathcal{M}_{n,n}(\mathbb{R})$ a positive definite matrix, $b \in \mathbb{R}^n$ and $c \in \mathbb{R}. $
$f : \mathbb{R}^n \to \mathbb{R}$ defined by : $$f(x) = \frac{1}{2}\langle Ax, x \rangle + \langle b, x \rangle + c$$
How do I bound this from below?
So far, here is what I have:
$f(0) = c$
$\lvert \lvert x\rvert \rvert \geq 0$
By Cauchy Schwartz, $-\lvert \lvert b\rvert \rvert \ \lvert \lvert x\rvert \rvert \leq \langle b, x \rangle \leq \lvert \lvert b\rvert \rvert \ \lvert \lvert x\rvert \rvert $
And by positive definiteness of $A, \exists \sigma \in (0, + \infty)$ such that $\frac{1}{2}\langle Ax, x \rangle \geq \sigma' \lvert \lvert x\rvert \rvert^2$
Also, we know that $f$ is coercive ie $f \to +\infty$ as $\lvert \lvert x\rvert \rvert^2 \to +\infty$
Giving us: $f(x) \geq \sigma' \lvert \lvert x\rvert \rvert^2 - \lvert \lvert b\rvert \rvert \ \lvert \lvert x\rvert \rvert +c \geq - \lvert \lvert b\rvert \rvert \ \lvert \lvert x\rvert \rvert +c $
Am I on the right track or is there another approach? Is there something obvious I have missed?
