The necessary part of the proof is really easy:

If $x^TAx$ can have negative values, then $f$ has no minimum on $\Bbb R^n$, because the term $ c^Tx$ is linear in $x$ and $x^TAx$ is bilinear.

Hence, we need $A$ to be positive semidefinite.

If $A$ is positive definite, then everything is ok, we can even find $x_\ast$ where minimum is attained: $x_\ast = -\frac 12 A^{-1}c$.

Suppose that $\ker A$ is non-zero, then take $x\ne 0$, $x\in \ker A$. $f(x)=c^Tx$. This is a linear function on $\ker L$, hence it can not have a minimum, unless it's constantly zero.

Therefore, we conclude the necessary conditions: $A$ is positive semidefinite and $\forall x\in \ker A$ we have $c^Tx=0$.

Let's now prove that these conditions are, in fact, sufficient. This part of the proof is slightly more difficult. Define $U=\ker A$ and $V=(\ker A)^\bot$. Clearly, $U\oplus V=\Bbb R^n$.

Let us also define $\lambda$ - smallest nonzero eigenvalue of $A$. Clearly, $\forall v\in V$ we have $v^TAv\ge \lambda |v|^2$ (ask if unclear why).

Furthermore, let $x=u+v$ such that $u\in U$ and $v\in V$ (this decomposition is unique). We can show that there exists $r>0$ such that $f(x)\ge 0$ for all $x=u+v$ such that $|v|> r$. Indeed, then $$f(x)=v^TAv + c^Tv\ge \lambda |v|^2+ c^Tv\ge |v|(\lambda|v|-|c|).$$ The last expression is non-negative when $|v|\ge \frac{|c|}{\lambda}$. We can now take $r =\frac{|c|}{\lambda} $.

The above result shows that on the set $E =\{(u+v):u\in \ker A,\;v\in (\ker A)^\bot, |v|\ge r\}$ our function $f$ is non-negative, hence $E$ in not interesting from minimization point of view (we know that $f(0)=0$). We now need to examine the complementary $E^c =\{(u+v):u\in \ker A,\;v\in (\ker A)^\bot, |v|\le r\} $.

We write
$$0\ge \inf_{x\in \Bbb R^n} f(x)=\inf_{x\in E^c} f(x) = \inf_{v\in (\ker A)^\bot, |v|\le r}v^TAv + c^Tv $$
(we used that $c\in (\ker A)^\bot$). The last infimum exists and is finite because we minimize a continuous function on a compact set. Therefore, the sufficient part is also proven.

As for geometric understanding, I'd try to use convexity, but I'm not really apt to produce such proof.

## Best Answer

Here is a solution with no calculus at all. It's the same process of completing the square that we learn in the first algebra course in high school with some linear algebra thrown in.

Let $q=\frac12 A^{-1}b$. Note that $$f(x) = (x+q)^\top A (x+q) + (c-q^\top Aq).$$ Notice that we use symmetry of $A$ to get $q^\top Ax= \frac12 b^\top (A^{-1}Ax) = \frac12 b^\top x$ and similarly for $x^\top Aq$. Since $A$ is positive definite, $y^\top Ay\ge 0$, with equality holding if and only if $y=0$. Thus, $f$ attains its minimum when $x+q=0$, i.e., when $x=-q=-\frac12 A^{-1}b$.