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

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.

• Following the hint in comment: apply the inequality AM$\ge$ GM on $x_1y_2+x_2y_1$.
– Koro
May 8, 2021 at 20:40
• The $\{\text{curly braces}\}$ belong within the MathJax code, not outside of it, thus: $S = \{(x_1,x_2):x_1x_2\geq4,x_1>0,x_2>0\}$ $$S = \{(x_1,x_2):x_1x_2\geq4,x_1>0,x_2>0\}$$ I edited this question accordingly. $\qquad$ May 8, 2021 at 20:57

\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}
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 $$S$$.