Show that the set $\{x \in \mathbb{R}_+^3 \mid x_1 x_2 x_3 \geq 1 \} $ is convex

Question

I found a similar question, but I do not understand how I can solve mine:

Show that the set $\{x \in \mathbb{R}_+^3 \mid x_1 x_2 x_3 \geq 1 \} $ is convex.

Can you help me with this?

As a starting point: can you show that if $(x_1,x_2,x_3)$ and $(y_1,y_2,y_3)$ are both in the set, then their midpoint is also in the set? — Greg Martin, Apr 04 '18 at 07:57

score 1 · Answer 1 · edited Apr 04 '18 at 10:16

1

Your set is the epigraph of the function $f(x,y)=\frac{1}{xy}$. Use the fact that $f$ is convex $\iff$ its Hessian is positive semidefinite, and the claim follow.

edited Apr 04 '18 at 10:16

Rodrigo de Azevedo

1

answered Apr 04 '18 at 08:05

Surb

55,662

score 1 · Accepted Answer · answered Apr 04 '18 at 10:09

1

Let $S=\{x\in\mathbb{R}^3 : x_1x_2x_3 \geq 1\}$ with $x=(x_1, x_2, x_3), y=(y_1, y_2, y_3) \in S$. For all $t \in (0, 1)$, by Young's inequality we have

$tx_i+(1-t)y_i \geq x_i^ty_i^{1-t}$ for $i=1,2,3$; thus $\prod_{i=1}^{3}(tx_i+(1-t)y_i) \geq (\prod_{i=1}^{3}x_i)^{t}(\prod_{i=1}^{3}y_i)^{1-t} \geq 1$ and $tx+(1-t)y \in S$, as desired.

answered Apr 04 '18 at 10:09

shrimpabcdefg

1,524

Exactly what I wrote yesterday ! Thank you :) – Sim Apr 05 '18 at 03:49

Show that the set $\{x \in \mathbb{R}_+^3 \mid x_1 x_2 x_3 \geq 1 \} $ is convex

2 Answers2