Let $A$ be a positive definite, symetrical matrix.
Is it true, that $A\cdot x \leq b$ $\Leftrightarrow$ $x \leq A^{-1}\cdot b$.
It is clear, that if $A$ is positive definite, than $A$ must be invertible as all the eigenvalues are positive in that case.
I just found an counterexample:
Let $A = \pmatrix{2 & -1 \\ -1 & 1}$, $x = \pmatrix{-1 \\ 0}$ and $b = \pmatrix{0 \\ 0}$.
$A$ is positive definite, as both eigenvalues are positive. $A\cdot x = \pmatrix{-2 \\ 1}$, so not less or equal $b$, however $ x \leq A^{-1} \cdot b$
