Properties of Nonnegative Orthant Let 
$$\mathbb{R}_+^n = \{x = (x_1,...,x_n)| x_1 \geq 0, x_2 \geq 0,\cdots, x_n \geq 0 \}$$ 
be the nonnegative orthant of $\mathbb{R}^n$.

How can we demonstrate the properties of the monotonic cone?

1) Let $c \in \mathbb{R}^n$ be given. Show that $c^Tx \geq 0 \forall x \in \mathbb{R}_+^n$ if and only if $c \in \mathbb{R}_+^n$.

2) Let $x, y \in \mathbb{R}_+^n$. Show that $x^Ty=0$ if and only if $x_iy_i = 0$ for each $i=1,...n$

First, I was thinking if we show the monotone cone is self-dual then $c \geq 0$, and therefore $c \in \mathbb{R}_+^n$.

Another part of it, suppose if  $c \leq 0$, then $c$ not $\mathbb{R}_+^n$, and $c^Tx \leq 0$. But this is a contradiction.
Hints

1) One direction is obvious. For the other direction note that the canonical basis is nonnegative, that is $e^i \in \Bbb R_n^+$ for every $i$, where $(e^i)_j:=1$ if $i=j$ and $(e^i)_j:=0$ otherwise.

2) Again, one direction is obvious. For the other direction, suppose by contradiction that $x^\top y=0$ and $x_iy_i \neq 0$, then $x_iy_i&gt;0$ as $x,y\in\Bbb R^n_+$ and so $x^\top y\geq x_iy_i&gt;0$, a contradiction.
Hints

1) One direction is obvious. For the other direction note that the canonical basis is nonnegative, that is $e^i \in \Bbb R_n^+$ for every $i$, where $(e^i)_j:=1$ if $i=j$ and $(e^i)_j:=0$ otherwise.

2) Again, one direction is obvious. For the other direction, suppose by contradiction that $x^\top y=0$ and $x_iy_i \neq 0$, then $x_iy_i&gt;0$ as $x,y\in\Bbb R^n_+$ and so $x^\top y\geq x_iy_i&gt;0$, a contradiction.