Solution
= Solution
Because $A$ is <Hermitian positive-definite matrix>[symmetric positive definite],
$$
\nabla f(x)=Ax-b,\qquad \nabla^2f(x)=A\succ0.
$$
Thus the only stationary point is $x_*=A^{-1}b$, and
$$
f(x)=f(x_*)+\frac12(x-x_*)^TA(x-x_*).
$$
The final term is positive unless $x=x_*$, proving that $A^{-1}b$ is the unique minimizer.