$g$ is coercive for $g(x)=x^TAx+b^Tx+c$ Suppose $A$ is a symmetric positive definite matrix $A\in \Bbb{R}^{n\times n}$, $b \in \Bbb{R}^n$, and c is a real number. Let $$g(x)=x^TAx + b^Tx + c$$
Show that $g$ is coercive.
Because $A$ is positive definite, then $x^TAx\gt0\ \ , \forall x \ne0$
I tried expanding out $g(x)$ and got $$g(x)=(a_{11}x_1^2+a_{22}x_2^2+...+a_{nn}x_n^2+2x_1(a_{12}x_2+a_{13}x_3+...+a_{1n}x_n)+2x_2(a_{23}x_3+...+a_{2n}x_n)+...+2a_{n-1,n}x_{n-1}x_n)+(b_1x_1+b_2x_2+...+b_nx_n)+c$$
Now I'm stuck as to how to show that $$\lim_{|x|\to\infty} g(x)=\infty$$
 A: As $A$ is positive definite, let $C>0$ so that $x^T Ax\ge C|x|^2$ for all $x$. Then 
$$|g(x)| = |x^TA x + bx + c| \ge C|x|^2 - |b| \cdot |x| - |c| \to \infty$$
as $|x| \to \infty$. 
A: Recall that  $\,g:\mathbb R^n \to \mathbb R\,$ is coercive if for any $\, x\in \,\mathbb R^n\,$ 
$%there exists a constant \,c\in \mathbb R\, such that$ 
$
%\big\langle \,g\left(x\right),\,x\,\big\rangle  \ge c\, \left\| \, x \, \right\|^2 
%= c \,\big\langle \,x,\,x\,\big\rangle
\left\| \, x \, \right\| \to \infty \implies  
\left\lvert \,g\left(x\right)\,\right\rvert  \to +\infty.$
Now, in your case $\,g\left(x\right)=x^TAx + b^Tx + c\,$ so that
\begin{align}
\big\lvert \,g\left(x\right)\big\rvert  
& = \left\lvert \,x^TAx + b^Tx + c\,\right\rvert  
\\ & \geq  
\left\lvert\,x^TAx\,\right\rvert-\left\lvert\,b^Tx\,\right\rvert - \left\lvert \,c\,\right\rvert  
\\ & =
%\left\lvert\,x^TAx\,\right\rvert-\left\|\,b^Tx\,\right\| - \left\lvert\,c\,\right\rvert  
\big\langle \,x,\, Ax\,\big\rangle - \big\langle \,b,\, x\,\big\rangle - \left\lvert \,c\,\right\rvert  
\\ & \ge 
a\, \left\| \,x\,\right\|^2 - \left\| \,b \,\right\|_\infty \left\| \,x\,\right\| - \left\lvert \,c\,\right\rvert  ,
\end{align}
where 
$\displaystyle\, a:= \min_{i,j=1,\dots, n} \left\lvert\, A_{ij} \,\right\rvert $
is the minimum absolute value of entries of matrix $\,A$.
Therefore we conclude that  $\, g\,$ is coercive:
$$
\lim_{\left\|x\right\| \to \infty} \big\lvert \,g\left(x\right)\big\rvert  
=\lim_{\left\|x\right\| \to \infty}  \left(
a \left\| \,x\,\right\|^2 - \left\| \,b \,\right\|_\infty \left\| \,x\,\right\| - \left\lvert \,c\,\right\rvert  \right)
= + \infty,
$$

EDIT:
Thanks to comment by John MA I correct a mistake and elaborate on the last inequality. 
Let
$$
A = 
\begin{bmatrix}
a_{11} & a_{12} & \cdots & a_{1n} \\
a_{21} & a_{22} & \cdots & a_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
a_{n1} & a_{n2} & \cdots & a_{nn} \\
\end{bmatrix},
\quad x = 
\begin{bmatrix} x_{1} \\ x_{2} \\ \vdots \\ x_{n}\end{bmatrix},
$$
so that
$$
Ax = 
\begin{bmatrix} 
\displaystyle\sum_{i=1}^n a_{1i} \,x_{i} \\ 
\displaystyle\sum_{i=1}^n a_{2i} \,x_{i} \\ 
\vdots \\ 
\displaystyle\sum_{i=1}^n a_{ni} \,x_{i}
\end{bmatrix},
$$
Denote $\displaystyle\, a:= \min_{a_{ij}\in A} a_{ij}\ge 0\,$ (since $\,A\,$ is symmetric positive-definite), then
\begin{align}
\big\langle \,x,\, Ax\,\big\rangle 
&= 
x_1 \sum_{j=1}^n a_{1j} \,x_{j}  + 
x_2 \sum_{j=1}^n a_{2j} \,x_{j} + 
\cdots +
x_n \sum_{j=1}^n a_{nj} \,x_{j}
\\ &=
x_1^2 \sum_{i=1}^n a_{i1}  + 
x_2^2 \sum_{i=1}^n a_{i2}  + 
\cdots
x_n^2 \sum_{i=1}^n a_{i1}  + 
\sum_{\substack{i,j=1\\i\neq j}}^n a_{ij} \,x_i \, x_j 
\\ & = 
\sum_{j=1}^n \,  x_j^2 \, \sum_{i=1}^n a_{ij} + 
\sum_{\substack{i,j=1\\i\neq j}}^n a_{ij} \,x_i \, x_j 
\\ &\ge \sum_{j=1}^n\,x_j^2 \,\sum_{i=1}^n a_{ij} =a\cdot n\sum_{j=1}^n x_j^2 
\\ & = 
a\cdot n\cdot \left\|\,x\, \right\|^2
\end{align}
Therefore we conclude that 
\begin{align}
\big\lvert \,g\left(x\right)\big\rvert  
& \ge 
\big\langle \,x,\, Ax\,\big\rangle - \big\langle \,b,\, x\,\big\rangle - \left\lvert \,c\,\right\rvert  
\\ & \ge 
a\, \left\| \,x\,\right\|^2 - \left\| \,b \,\right\|_\infty \left\| \,x\,\right\| - \left\lvert \,c\,\right\rvert,
\end{align}
and therefore $\,g\left(x\right)\,$ is coercive.
