How to check whether a given inequality is convex?

Question

I have the inequality

$$x_1^2+x_2^2\geqslant1$$

How do I check its convexity analytically?

Answer 1

We have that $x_1^2+x_2^2=1$ gives the zero locus of a circle of radius $1$. $x_1^2+x_2^2=r^2$ gives a circle of radius $r$. You are taking the set in $\Bbb R^2$ consisting of all points lying on the circles of radius $[1,\infty)$. Take the convex hull of this, is it the same thing. Take the convex hull of any point laying on the locus of $x_1^2+x_2^2=1$.

Also, consider the open ball of radius $1$ centered on the origin, this is the complement of your set.

Answer 2

$(1,0)$ and $(0,1)$ are in the set, but $(0,0)$ is on the straight line which joins the points and not in the set. So the set isn't convex.