prove a set that the coordination multiplication is bigger than a certain value to be convex set

Question

I have a set here $$\{(x,y)|xy \geq c, x>0, y>0\}$$ how to prove this set is a convex set for any c?

Namely, how to prove $$(\alpha x_1+(1-\alpha)x_2)(\alpha y_1+(1-\alpha)y_2)\geq c, \forall 1>\alpha>0$$ if $x_1y_1\geq c$ and $x_2y_2\geq c$

Answer 1

Take $c>0$ (otherwise it's obvious). Now \begin{align} (\alpha x_1+(1-\alpha)x_2)(\alpha y_1+(1-\alpha)y_2) &= \alpha^2x_1y_1 + \alpha(1-\alpha)x_2y_1 + \alpha(1-\alpha)x_1y_2 + (1-\alpha)^2x_2y_2 \\ &\ge(\alpha^2+(1-\alpha)^2)c + \alpha(1-\alpha)(x_2y_1+x_1y_2). \end{align} By AM-GM, we have $$x_2y_1+x_1y_2\ge 2\sqrt{x_2y_1x_1y_2} \ge 2c $$ and hence $$ (\alpha x_1+(1-\alpha)x_2)(\alpha y_1+(1-\alpha)y_2) \ge c(\alpha^2+(1-\alpha)^2+2\alpha(1-\alpha)) = c(\alpha+(1-\alpha))^2 = c$$ as desired.

Answer 2

Directly from the definition. Take two points $(x,y)$ and $(x',y')$ with $xy \ge c, x'y'\ge c$. Parameterize the line segment connecting them and show that for all points $(a,b)$ on the segment, $ab \ge c$. You really don't care about the value of $c$ because the proof goes through if you divide $c$ and all the $x$ values (alternately all the $y$ values) by $c$. The argument changes a bit if $c \le 0$