Prove that $1- x^T Qx- b^Tx$ is not convex

Question

$Q$ is positive definite ($Q>0$). I want to prove that problem $P$ is not convex. How can I do it?

Hint: Complete the square.

Why can't I say that $1-x^TQx-b^Tx \leq 0$ is a concave function?

Answer 1

Suppose that $f(x)=1-x^TQx-b^Tx$ is convex. Then $f(\frac{x+y}{2}) \le \frac{f(x)+f(y)}{2}$ for all $x,y$. For $y=0$ this gives:

$x^TQx \le -2$, which is absurd !