# Sum of positive elements divided by their “weighted” product - inequality

I have following expression,

$$\frac{\sum_{i=1}^n x_i}{\prod_{i=1}^nx_i^{p_i}}$$

where $$p_i$$s satisfy $$\sum p_i = 1$$ and $$p_i \in [0,1]$$ and $$x_i\geq0$$, $$\forall i \in 1\dots n$$.

I think that this expression is always $$\geq 1$$, however, I don't know how to prove it.

Is there anything I can conclude?

Thanks.

• The weighted AM-GM inequality yields $\prod_i x_i^{p_i} \leq \left(\sum_i p_i x_i\right)^{\sum_i p_i} = \left(\sum_i p_i x_i\right)^1 = \sum_i p_i x_i \leq \sum_i 1 x_i = \sum_i x_i$. – darij grinberg Jan 10 '19 at 12:40
• @darijgrinberg That comment could be made an answer. [Anyway I'd upvote it.] – coffeemath Jan 10 '19 at 12:45
• @coffeemath: Good idea. – darij grinberg Jan 10 '19 at 12:49

Applying the weighted AM-GM inequality to the weights $$p_1, p_2, \ldots, p_n$$, we obtain $$$$\dfrac{p_1 x_1 + p_2 x_2 + \cdots + p_n x_n}{1} \geq \sqrt[1]{x_1^{p_1} x_2^{p_2} \cdots x_n^{p_n}}$$$$ (since $$p_1 + p_2 + \cdots + p_n = \sum_{i=1}^n p_i = 1$$). This simplifies to $$$$p_1 x_1 + p_2 x_2 + \cdots + p_n x_n \geq x_1^{p_1} x_2^{p_2} \cdots x_n^{p_n} .$$$$ Hence, \begin{align} x_1^{p_1} x_2^{p_2} \cdots x_n^{p_n} \leq p_1 x_1 + p_2 x_2 + \cdots + p_n x_n = \sum_{i=1}^n \underbrace{p_i}_{\substack{\leq 1 \\ \text{(since p_i \in \left[0,1\right])}}} x_i \leq \sum_{i=1}^n x_i , \end{align} so that \begin{align} \sum_{i=1}^n x_i \leq x_1^{p_1} x_2^{p_2} \cdots x_n^{p_n} = \prod_{i=1}^n x_i^{p_i} . \end{align}
Concavity of $$\log$$ gives $$\log\left( \prod_{i=1}^nx_i^{p_i}\right) = \sum_{i=1}^n p_i\log x_i \stackrel{\mbox{concavity}}{\leq} \log\left( \sum_{i=1}^np_i x_i\right) \stackrel{0\leq p_i\leq 1}{\leq} \log\left( \sum_{i=1}^nx_i\right)$$