Is this function coercive? We are given $A \in \mathcal{M}_{n,n}(\mathbb{R})$ a positive definite matrix, $b \in \mathbb{R}^n$ and $c \in \mathbb{R}. $
Our function is $f : \mathbb{R}^n \to \mathbb{R}$ defined by : 
$$f(x) = \frac{1}{2}\langle Ax, x \rangle + \langle b, x \rangle + c$$
A function is said to be coercive if: 
$$f(x) \to +\infty \text{ as } \lvert \lvert x \rvert \rvert _{2} \to +\infty.$$
I normally attempt to bound the function from below by some function that is also coercive but am unable to do so here. 

By applying the Spectral Theorem, we can write $A = \sum_{i =1}^k \lambda_i P_{\lambda_i}$, where $\lambda_i$ is the $i$-th distinct eigenvalue of $A$ and $P_{\lambda_i}$ is the orthogonal projection matrix onto the eigenspace of $\lambda_i$.  Since $A$ is positive defininite, each $\lambda_i >0$.  If we set $\lambda$ to be the smallest eigenvalue of $A$, then $\langle Ax,x \rangle = \sum_{i=1}^k \lambda_i\|P_{\lambda_i} x\|_2^2 \geq \lambda \sum_{i=1}^k \|P_{\lambda_i} x\|_2^2 = \lambda \|x\|_2^2$.   By Cauchy-Shwarz, $\langle b, x\rangle \geq -\|b\|_2\|x\|_2$.  Thus we get the lower bound $f(x) \geq \frac{\lambda}{2}\|x\|_2^2 -\|b\|_2\|x\|_2 + c$, which goes to infinity with $\|x\|_2$.
