Is $2x + y \geq k$ a convex set?

Question

I'm asked to prove that $2x+y \geq k$ is a convex set.

My teacher told me to take two points $(x_1,y_1)$ and $(x_2,y_2)$ within the set.

Then he showed two inequalities:

1. $2 x_1 + y_1 \geq k$
2. $2 x_2 + y_2 \geq k$

After that he multiplied the inequalities by $\lambda$ and $[1-\lambda]$

3. $(2 x_1 + y_1 ) \lambda \geq 6 \lambda$
4. $2 x_2 + y_2 [1-\lambda] \geq 6[1- \lambda]$

And he ended up with this expression: $2 x_1 \lambda +2 x_2 [1-\lambda]+ y_1 \lambda + y_2 [1-\lambda] \geq k$ to finally proof the set is convex.

My doubt is how he manage to get from $3$ and $4$ to the last expression. Thanks to anyone who take the time to help me.

Answer 1

Let $f(x, y) = 2x + y$, since $f$ is linear it is both concave and convex. The set $\{(x, y) : 2x + y \geq k\} = \{(x, y): f(x, y) \geq k\}$ is a superlevel set of $f$. Since $f$ is concave, this set is convex.