why can we use this $x^\theta y^{1-\theta} \le \theta x+(1-\theta)y$ to prove the $\prod \limits_{i=1}^{n}x_i \ge 1$ is convex set?

Question

Show that $\{x \in R^n_+|\prod \limits_{i=1}^{n}x_i \ge 1\}$ is convex set

Hint : if $x ,y \ge 0$ and $0\le \theta \le 1$,then $x^\theta y^{1-\theta} \le \theta x+(1-\theta)y$

I don't understand the relation between the problem and the hint,the question wants me to prove the $\prod \limits_{i=1}^{n}x_i \ge 1$ is a convex,why shouldn't we use the definition of convex: $f(\theta x+(1-\theta )y) \le \theta f(x)+(1-\theta)f(y)$,but use this $x^\theta y^{1-\theta} \le \theta x+(1-\theta)y$ to prove the $\prod \limits_{i=1}^{n}x_i \ge 1$ is convex set?

Answer 1

you may want to use this hint the other way: from right to left...

let's call $\mathbb{A}$ your ensemble, let's take $(x,y) \in \mathbb{A}^2$. to prove that $\mathbb{A}$ is convex, we need to prove that $\forall t \in [0,1], z = tx+(1-t)y \;$ is in $\mathbb{A}$

$\Pi_{i=1}^n z_i = \Pi_{i=1}^n (tx_i+(1-t)y_i)$

now you use the inequality from the right side to the left.

$\Pi_{i=1}^n (tx_i+(1-t)y_i) > \Pi_{i=1}^n x_i^ty_i^{1-t}$

you are now one line away from the result