Show that $\{(x_1,x_2):x_1x_2\geq4,x_1>0,x_2>0\}$ is a convex set

Question

Trying to show that the set $S = \{(x_1,x_2):x_1x_2\geq4,x_1>0,x_2>0\}$ is convex.

I tried proving by definition, let $x=(x_1,x_2)$ and $y=(y_1,y_2)$ and $\lambda\in[0,1]$ We want to show that $\lambda x + (1-\lambda)y \in S$

for $\lambda=1$ and $\lambda=0$ the answer is trivial. for $\lambda \in (0,1)$ I got a bit stuck

What I tried was

$\lambda x + (1-\lambda)y = \lambda(x_1,x_2) + (1 - \lambda)(y_1,y_2) = \lambda(x_1,x_2) - \lambda(y_1,y_2) + (y_1,y_2) = (\lambda x_1 - \lambda y_1 + y_1, \lambda x_2 - \lambda y_2 + y_2)$

So now if we can prove that $(\lambda x_1 - \lambda y_1 + y_1)\times(\lambda x_2 - \lambda y_2 + \lambda y_2) \geq 4$ we are done.

I got stuck here, I tried looking for a way with the mean inequality but with no success. I think I am way off base, perhaps Jensen's inequality or something in that domain might be what I am looking for.

Answer 1

Hint:

$\begin{align}(\lambda x_1+(1-\lambda)y_1)(\lambda x_2+(1-\lambda)y_2)=&\lambda^2x_1x_2+\lambda(1-\lambda)(x_1y_2+x_2y_1)+(1-\lambda)^2y_1y_2\\&\ge \lambda^2x_1x_2+2\lambda(1-\lambda)\sqrt {x_1x_2y_1y_2 }+(1-\lambda)^2y_1y_2 \end{align}$

Can you take it from here?

Answer 2

Consider the graph of $y = \frac{4}{x}$ for $x > 0$. Note that the region including and above this graph is your set $S$. That is, the set of points $(x, y)$ such that $xy \ge 4$ and $x > 0$. The function $f(x) = \frac{4}{x}$ is convex for $x > 0$ since its second derivative is strictly positive. It is known that a function is convex if and only if its epigraph is convex. The epigraph of $f$ in this case is exactly $S$.