Positivity of a function in $\mathbb{R}^{n}$ We place ourself in $\mathbb{R}^{n}$. We consider a given increasing function 
$$ g : \begin{aligned}
&\mathbb{R}^{+} \to \mathbb{R}
\\
&x \;\;\,\mapsto g(x)
\end{aligned}$$
Finally, we introduce the function $F (x_{i})$ defined as
$$ F : 
\begin{aligned}
&(\mathbb{R}^{+})^{n} \to \;\; \mathbb{R}
\\
&(x_{i}) \;\;\;\mapsto \;\; n \left[ \sum\limits_{i = 1}^{n} x_{i} \, g (x_{i}) \right] - \left[ \sum\limits_{i = 1}^{n} x_{i} \right] \left[ \sum\limits_{i = 1}^{n} g(x_{i}) \right]
\end{aligned}
$$
Note that $F$ is only defined for sets of positive coefficients.
My question is :
Is it true that $\forall n \geq 1, \forall (x_{i}) \in (\mathbb{R}^{+})^{n} \;,\; F (x_{i}) \geq 0 $ ? Or do I need to add other constraints on the $g$ function ?
For example, in the case $n=2$, one has
$2 (x_{1} g (x_{1}) + x_{2} g (x_{2})) - (x_{1} + x_{2}) (g(x_{1}) + g (x_{2})) = (x_{1} - x_{2}) (g (x_{1}) - g (x_{2}))$
Such a term is positive because we have supposed that $g$ is increasing.
I tried various approaches : Cauchy-Schwarz Inequality, norms $\Vert \cdot \Vert_{1}$ and $\Vert \cdot \Vert_{2}$ comparison, and Riemann sum formula, but I am unfortunalety unable for the moment to handle the general case. Maybe looking at $\frac{\partial F}{\partial \boldsymbol{x}}$ ?
 A: This is true with no extra conditions, it follows from a variation on the rearrangement inequality, called  Chebyshev's inequality
(see http://en.wikipedia.org/wiki/Rearrangement_inequality.): If $a_1\leq a_2\leq\ldots\leq a_n$ and 
$b_1\leq b_2\leq\ldots\leq b_n$, then
$$
\frac{1}{n}\sum_{i=1}^na_ib_i\geq\left(\frac{1}{n}\sum_{i=1}^na_i\right)\left(\frac{1}{n}\sum_{i=1}^nb_i\right)
$$
we just reorder increasingly the $x_i$'s to get the $a_i$'s and we set $b_i=g(a_i)$. 
A: Without loss of generality $x_1\le x_2\le\ldots\le x_n$, then $g(x_1)\le g(x_2)\le\ldots\le g(x_n)$. So by the rearrangement inequality it's true that
$$\begin{align}x_1g(x_1)+x_2g(x_2)+\ldots+x_ng(x_n)&\ge x_1g(x_1)+x_2g(x_2)+\ldots+x_ng(x_n)\\
x_1g(x_1)+x_2g(x_2)+\ldots+x_ng(x_n)&\ge x_2g(x_1)+x_3g(x_2)+\ldots+x_1g(x_n)\\
&\ \vdots\\
x_1g(x_1)+x_2g(x_2)+\ldots+x_ng(x_n)&\ge x_ng(x_1)+x_1g(x_2)+\ldots+x_{n-1}g(x_n)\end{align}$$
Adding all these inequalities we get
$$n\left( \sum\limits_{i = 1}^{n} x_{i} \, g (x_{i}) \right) \ge \left( \sum\limits_{i = 1}^{n} x_{i} \right) \left( \sum\limits_{i = 1}^{n} g(x_{i}) \right)$$
Note that you don't even need $x_i\in\mathbb R^+$, it holds for arbitrary $x_i\in\mathbb R$.
