Positivity of a function in $\mathbb{R}^{n}$

Question

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 $\frac{\partial F}{\partial \boldsymbol{x}}$ ?

Answer 1

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 $x_i\in\mathbb R$.

Answer 2

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 $b_i=g(a_i)$.