In exercise 5 of section 0 of Fundamentals of convex analysis by Hiriart-Urrut, Lemaréchal, we're supposed to prove that if a self-adjoint linear operator $A:\mathbb{R}^n \rightarrow \mathbb{R}^n$ is positive semi-definite and $b \in \textrm{Im}A$, then $q(x)=\langle Ax,x \rangle - 2 \langle b,x \rangle$ has a finite infimum, and this infimum is a minimum.
I'm not sure how to prove this. An outline of a possible approach:
We will show that $q$ grows large as we move away far away from the origin. If that is true, then for a properly chosen $U=\overline{B}(0,K)$ we have that $q$ is continuous and so it achieves a minimum on $U$. Because $q$ only "grows" outside of $U$, this minimum is global.
Denote the symmetric bilinear form $(x,y) \mapsto \langle Ax,y \rangle$ by $L$. Then it's easy to see that if $L(v,v)>0$, then $L(kv,kv)=k^2 L(v,v)$, and so $q(kv)$ goes to infinity as we raise $k>0$. By the spectral theorem there exists a basis $\{ b_1,...,b_n \}$ such that $L$ with respect to this basis is diagonal and the elements on the diagonal are non-negative.
If the elements on the diagonal are all positive we are done by the previous part. If the $i$th element on the diagonal is zero, then $q(b_i)=L(b_i,b_i)-2 \langle b,x \rangle = L(b_i,b_i) - 2 \langle Ay, b_i \rangle = L(b_i,b_i) - 2L(b_i,y)=0-0=0$ (using the fact that $b \in \textrm{Im} A$).
However this seems excessive and I doubt I'm supposed to use the spectral theorem, so I think I must be missing something. EDIT: spectral theorem might not be necessary - symmetric elementary operations are enough to transform $L$ into a diagonal bilinear form with respect to some basis.