Given $$ A = \begin{bmatrix} 4 & 0 & 1 \\ 0 & 2 & 0 \\ 1 & 0 & 1 \end{bmatrix}, \qquad B = \begin{bmatrix} 1 & 2 & -3 \\ 5 & 2 & 1 \end{bmatrix}, \qquad b = \begin{bmatrix} 0\\ 10\end{bmatrix} $$
minimize $x^T A x$ subject to $B x \leq b$.
I have the answer which is $(0, 0, 0)$. However, my professor says the problem can be solved without doing any work. I've looked at this for a long amount of time and can't figure out why that would be the case, however I suspect it may have something to do with the fact $A$ is symmetric. Can someone please help me out and let me know how this can be solved without doing any work?
$A$ is a positive definitive bilinear form so $x^TAx > 0$ for all $x \neq 0$. As $0$ satisfies your equation, it must be the answer.