I have that $A$ is a positive definite matrix and $b$ is some fixed vector. Define $r=b-Ax$ to be the residual vector and $e=A^{-1}b-x$ to be the error vector for all vectors $x$. I need to show that $\langle e,r\rangle$ is positive unless $Ax=b$. What I have done so far:
\begin{align*} \langle e,r\rangle&=(A^{-1}b-x)^t(b-Ax)\\ &=(b^tA^{-1}-x^t)(b-Ax)\\ &=b^tA^{-1}b-b^tx-x^tb+x^tAx\\ &=b^tA^{-1}b+x^tAx-2x^tb. \end{align*}
I know that $b^tA^{-1}b$ and $x^tAx$ must be positive since $A$ (and $A^{-1}$ in turn) is positive definite. My issue is I don't know how to show that the $2x^tb$ term does not exceed $b^tA^{-1}b+x^tAx$. Any ideas/hints?
Hint: Rewrite $A^{-1}b-x$ as $A^{-1}(b - Ax)$. (Or alternatively, rewrite $b-Ax$ as $A (A^{-1} b - x)$.)