How do I prove that this set is convex?

Question

How do I prove that this set is convex? $S_2 = \{(x_1, x_2 ) : x_1x_2 ≥ 4, x_1 > 0, x_2 > 0\}$

I failed to prove that with {tx1 + (1 − t)x2 : 0 ≤ t ≤ 1}

Answer 1

As you are in a 2-dimensional space it is not enough to test for $\{tx_1 + (1-t)x_2: 0\leq t\leq 1\}$. You need to proof that for each $(x_1,x_2),(y_1,y_2)\in S_2$ the convex hull of those two points lies inside $S_2$ as well. Let $t\in[0,1]$ be arbitrary and take a point from the convex hull:

$$ t\cdot(x_1,x_2) + (1-t)\cdot (y_1,y_2) = (tx_1+(1-t)y_1,tx_2+(1-t)y_2) =: (z_1,z_2) $$

1. $z_1>0$
   $tx_1 + (1-t)y_1 > t\cdot 0 + (1-t) \cdot 0 = 0$
2. $z_2>0$
   $tx_2 + (1-t)y_2 > t\cdot 0 + (1-t) \cdot 0 = 0$
3. $z_1z_2 \geq 4$: This is the part where a bit calculation is necessary:
   $$(tx_1+(1-t)y_1)\cdot (tx_2+(1-t)y_2) = t^2x_1x_2 + t(1-t)(x_1y_2+x_2y_1) + (1-t)^2y_1y_2)\\ \geq t^2\cdot 4 + (1-t)^2 \cdot 4 + t(1-t) \cdot \left(\frac{4y_2}{x_2} + \frac{4x_2}{y_2}\right) \\ \geq4\cdot (t^2 + (1-t)^2 + 2t(1-t)) = 4 \cdot(t + (1-t))^2 =4 $$ The first inequality comes from the fact that $x_1x_2, y_1y_2 \geq 4$ which can be rewritten as $x_1 \geq \frac{4}{x_2}$ and $y_1\geq \frac{4}{y_2}$. The second inequality arises from the simple equation $a^2>0$ together with $x_2,y_2>0$: $$ (x-y)^2 \geq 0 \Leftrightarrow x^2 + y^2 \geq 2xy \Leftrightarrow \frac{x}{y} + \frac{y}{x} \geq 2 $$