Let $$\Omega = \{(x_1,x_2)\in\mathbb{R}^2:x_1^2-x_2\leq 6\}$$
Prove that $\Omega$ is a convex set from first principles using the convex combination.
edit: Thanks Ewan for that, but I am trying to do it from the definition, i changed the question to state that now.
You have
$$\Omega = \{(x,y)\in\mathbb{R}^2:y \geq f(x)\}$$
where $f(x)=x^2-6$. Now $f$ is convex (because for example $f’’(x)=2 \geq 0$), so $\Omega$ is convex also.
If you want a derivative-free proof of the convexity of $f$ : for $\lambda \in [0,1]$ and $t_1,t_2 \in {\mathbb R}$,
$$ \lambda f(t_1) + (1-\lambda) f(t_2)- f(\lambda t_1+(1-\lambda) t_2)= \lambda (1-\lambda) ((t_1-t_2)^2) \geq 0 $$
If $M_1=(x_1,y_1)$ and $M_2=(x_2,y_2)$ are two points in $\Omega$, consider the convex combination $P=(1-\lambda)M_1+\lambda M_2=(x_3,y_3)$. Then
$$ x_3=\lambda x_1+(1-\lambda) x_2, \ y_3=\lambda y_1+(1-\lambda) y_2 $$
So
$$ y_3=\lambda y_1+(1-\lambda) y_2 \geq \lambda f(x_1)+(1-\lambda) f(x_2) \geq f(\lambda x_1+(1-\lambda) x_2)=f(x_3) $$
so $P\in\Omega$, and $\Omega$ is indeed convex.