Prove the existence and uniqueness of a global minimum for a quadratic function $q(x)$ with positive definite Hessian
Solution 1:
If $q(x)= \frac{1}{2} x^\top A x + b^\top x + c$ then you can "complete the square" to rewrite it as $$q(x) = \frac{1}{2} (x+A^{-1} b)^\top A (x+A^{-1} b) - \frac{1}{2} b^\top A^{-1} b + c.$$ Note that $A$ is invertible because it is positive definite. Moreover, the first term $(x+A^{-1} b)^\top A (x+A^{-1} b)$ is nonnegative because $A$ is positive definite. So, $$q(x) \ge -\frac{1}{2} b^\top A^{-1} b + c$$ for all $x$, and equality is clearly attained by the unique minimizer $x=A^{-1} b$.