[proofplan]
We show that $x^* = A^{-1}b$ is the unique global minimizer of $F(x) = \frac{1}{2}\langle x, Ax \rangle - \langle b, x \rangle$ by computing $F(x^* + h) - F(x^*)$ for an arbitrary perturbation $h$. The symmetry of $A$ yields a clean expansion in which the linear term vanishes (because $Ax^* = b$) and the quadratic term $\frac{1}{2}\langle h, Ah \rangle$ is strictly positive for $h \neq 0$ by positive definiteness. This establishes $x^*$ as the unique point where $F$ attains its global minimum.
[/proofplan]
[step:Expand $F(x^* + h) - F(x^*)$ using bilinearity and symmetry of $A$]
Let $h \in \mathbb{R}^n$ be arbitrary. Define $x^* := A^{-1}b$, so that $Ax^* = b$. Using the bilinearity of the inner product and the symmetry $A = A^\top$:
\begin{align*}
F(x^* + h) &= \frac{1}{2}\langle x^* + h, A(x^* + h) \rangle - \langle b, x^* + h \rangle \\
&= \frac{1}{2}\langle x^*, Ax^* \rangle + \frac{1}{2}\langle x^*, Ah \rangle + \frac{1}{2}\langle h, Ax^* \rangle + \frac{1}{2}\langle h, Ah \rangle - \langle b, x^* \rangle - \langle b, h \rangle.
\end{align*}
Since $A$ is symmetric, $\langle x^*, Ah \rangle = \langle Ax^*, h \rangle = \langle b, h \rangle$, so the two cross terms combine:
\begin{align*}
\frac{1}{2}\langle x^*, Ah \rangle + \frac{1}{2}\langle h, Ax^* \rangle = \frac{1}{2}\langle b, h \rangle + \frac{1}{2}\langle h, b \rangle = \langle b, h \rangle.
\end{align*}
Subtracting $F(x^*) = \frac{1}{2}\langle x^*, Ax^* \rangle - \langle b, x^* \rangle$:
\begin{align*}
F(x^* + h) - F(x^*) = \langle b, h \rangle + \frac{1}{2}\langle h, Ah \rangle - \langle b, h \rangle = \frac{1}{2}\langle h, Ah \rangle.
\end{align*}
[guided]
The strategy is to reduce the minimization problem to a simple positivity check. By expanding around the candidate minimizer $x^* = A^{-1}b$, we express the difference $F(x^* + h) - F(x^*)$ purely in terms of the perturbation $h$.
Using bilinearity of $\langle \cdot, A\cdot \rangle$:
\begin{align*}
F(x^* + h) &= \frac{1}{2}\langle x^*, Ax^* \rangle + \frac{1}{2}\langle x^*, Ah \rangle + \frac{1}{2}\langle h, Ax^* \rangle + \frac{1}{2}\langle h, Ah \rangle - \langle b, x^* \rangle - \langle b, h \rangle.
\end{align*}
The key simplification uses $A = A^\top$: the identity $\langle x^*, Ah \rangle = \langle A^\top x^*, h \rangle = \langle Ax^*, h \rangle = \langle b, h \rangle$ shows that the two cross terms combine to $\langle b, h \rangle$, which exactly cancels the $-\langle b, h \rangle$ from the linear part:
\begin{align*}
F(x^* + h) - F(x^*) = \underbrace{\langle b, h \rangle - \langle b, h \rangle}_{= 0} + \frac{1}{2}\langle h, Ah \rangle = \frac{1}{2}\langle h, Ah \rangle.
\end{align*}
The vanishing of the linear term in $h$ is precisely the statement that $\nabla F(x^*) = Ax^* - b = 0$, the first-order optimality condition. The entire difference is governed by the quadratic term $\frac{1}{2}\langle h, Ah \rangle$.
[/guided]
[/step]
[step:Conclude unique global minimality from positive definiteness of $A$]
Since $A$ is positive definite, $\langle h, Ah \rangle > 0$ for all $h \neq 0$. Therefore:
\begin{align*}
F(x^* + h) - F(x^*) = \frac{1}{2}\langle h, Ah \rangle > 0 \qquad \text{for all } h \neq 0.
\end{align*}
This means $F(x) > F(x^*)$ for every $x \neq x^*$, so $x^* = A^{-1}b$ is the unique global minimizer of $F$.
[/step]
