# Prove that the set $A := \left\{ (x,y) \in \Bbb R_{> 0}^2 \mid xy \geq 1 \right\}$ is convex [duplicate]

I tried taking a convex combination of $$x$$ and $$y$$ (given by $$(a,b)$$ and $$(s,t)$$ respectively) but the resulting expression had two terms which I could not get rid of or substitute with anything useful:

$$k(1-k)bs$$ $$k(1-k)at$$

I suspect I just have to manipulate the algebraic expression so that I can use the conditions $$ab >=1$$ and $$st >=1$$ but I haven't been able to do this.

Let, $$(x_1,y_1),(x_2,y_2)\in A$$

Again, let $$(x_3,y_3)=\lambda (x_1,y_1)+(1-\lambda)(x_2,y_2)$$ where $$0\le \lambda\le 1$$

Then, $$x_3y_3=\lambda^2x_1y_1+(1-\lambda)^2x_2y_2+\lambda(1-\lambda)(x_2y_1+x_1y_2)\ge \lambda^2+(1-\lambda)^2+2\lambda(1-\lambda)=1$$

and we are done!

• How do you show that $x_{2}y_{1} + x_{1}y_{2} >= 2$?
– John
Commented Sep 26, 2018 at 3:25
• Just using A.M.-G.M. inequality.
– SOUL
Commented Sep 26, 2018 at 3:28
• @Tom. can you clarify? I don't see the arithmetic mean here. Commented May 19, 2020 at 11:41
• ok, got it - $\frac{x_1 y_2 + x_2 y_1}{2} \ge \sqrt{x_1 y_2 x_2 y_1} \ge 1$ Commented May 19, 2020 at 12:06

An alternative is to use Proof that a function is convex if and only if its epigraph is convex.

Here since $$y>0$$ (in particular $$x\neq 0$$) the function $$y=f(x)=\frac 1x$$ is well defined and it is convex since $$f''(x)=\frac 2{x^3}>0$$ for $$x>0$$.