I'm new to both StackExchange and formal proofs, do let me know if there's any way I can improve my question-posing.
Given the following quadratic: $$f(x)=x^TAx+2b^Tx+c$$ $$A\succeq0 \ (A\in\Bbb R^{n\times n}), \ b\in\Bbb R^n, \ c\in\Bbb R$$ I'm attempting to show $f$ is bounded (uniformly, by a constant) below iff $b^T\in$ Range$(A)$. I get the feeling this is trivial and I'm missing something quite basic.
From my understanding, if $A\succ0 \implies \exists \ \alpha \ $ s.t. $f(x)\leq\alpha \; \forall x,b\in\Bbb R^n,$ without the need for any further conditions on $b$, and, by that same logic, as $A\succeq0$ the only instances in which we must constrain $b$ being when $x\in\left\{y:y^TAy=0, \; y\in\Bbb R^n\right\}$, as then for any $b\in\left\{d:d^Ty\neq0, \; y\in\Bbb R^n\right\} \; \nexists\alpha$ (the linear function will go to $-\infty$ in some direction).
Does the range encode this property, where if $x^TAx$ maps to zero, a $b^T \in$ Range$(A)$ maps $b^Tx$ to zero? It's unclear to me how; directly substituting in the range to the quadratic doesn't seem to elucidate things further:
$$x^TAx+2(Ay)^Tx+c, \; \forall y\in\Bbb R^n $$ $$=\left(x^T+2y^T \right)Ax+c$$
Does $x^TAx=0 \implies Ax=0 \; \left(x\in \ker(A)\right)$? That would at least afford us sufficiency.
As you said, the positive definite case is immediate and in the semidefinite case, we need to have $b^Tx=0$ for all $x\in\ker(A)$ (right null-space). That is $\ker(A)\subset\ker(b^T)$. It is more natural to me to consider the null-spaces to start-with.
You can then say that $b\bot\ker(A)$. Since the matrix is symmetric (so the row and column spaces are the same) and using the fact that the right-null space is orthogonal to the row-space, we get that the above statement is equivalent to $b\in\mathrm{Range}(A)$.
Regarding the last implication, this is correct. You can prove that using the fact that $x^TAx=\sum_{i=1}^r\lambda_i x^Tu_iu_i^Tx$ where $r$ is the rank of the matrix, $\lambda_i$ is the $i$-th eigenvalue and the $u_i$'s are the eigenvectors. Since the eigenvalues are positive and $x^Tu_iu_i^Tx\ge0$, then the sum is zero if and only if $u_i^Tx=0$. This implies that $Ax=0$.