Prove certain set is convex

Question

I'm having trouble proving this statement:

Prove that the set $$X=\{ (x,y)\in\mathbb{R}^2 : y\geq 0, x\leq b+y \}$$ given $b\in\mathbb{R}$ is convex.

The work I've done so far:

I first assumed we havo two different points $(x_1,y_1),(x_2,y_2)\in X$. Then, the segment between them is made of the points $$(\lambda x_1 + (1-\lambda)x_2, \lambda y_1 + (1-\lambda)y_2) \ , \ \lambda\in[0,1].$$ So if I prove that points are inside $X$ for any $\lambda$ I would have proven the convexity. It's easy to see that

$$\lambda y_1 + (1-\lambda)y_2 \geq 0,$$ since it's a positive number plus another positive number. Now, I don't know how to prove that

$$\lambda x_1 + (1-\lambda)x_2 \leq b + \lambda y_1 + (1-\lambda)y_2$$ knowing obviously that $x_1 \leq b+y_1$ and $x_2\leq b+y_2$. What can I do to prove this part? Any help or hint will be appreciated, thanks in advance.

Answer 1

Just multiply $x_1 \leq b+y_1$ by $\lambda$, $x_2 \leq b+y_2$ by $(1-\lambda)$ and add the inequalities. Finally note that $\lambda b +(1-\lambda) b=b$.

Answer 2

Let (x₁,y₁),(x₂,y₂) ∈ X Let (c,d) = (λx₁ + (1-λ)x₂, λy₁ + (1-λ)y₂) c = λx₁ + (1-λ)x₂ ≤ λ(b + y₁) + (1-λ)(b + y₂) = (λb + b - λb) + λy₁ + (1-λ)y₂ = b + d hence c ≤ b + d and you have already proved why d ≥ 0

It's actually pretty straight forward. You should've just expanded the expression.