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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.

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    $\begingroup$ The weighted AM-GM inequality yields $\prod_i x_i^{p_i} \leq \left(\sum_i p_i x_i\right)^{\sum_i p_i} = \left(\sum_i p_i x_i\right)^1 = \sum_i p_i x_i \leq \sum_i 1 x_i = \sum_i x_i$. $\endgroup$ – darij grinberg Jan 10 '19 at 12:40
  • $\begingroup$ @darijgrinberg That comment could be made an answer. [Anyway I'd upvote it.] $\endgroup$ – coffeemath Jan 10 '19 at 12:45
  • $\begingroup$ @coffeemath: Good idea. $\endgroup$ – darij grinberg Jan 10 '19 at 12:49
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Applying the weighted AM-GM inequality to the weights $p_1, p_2, \ldots, p_n$, we obtain \begin{equation} \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}} \end{equation} (since $p_1 + p_2 + \cdots + p_n = \sum_{i=1}^n p_i = 1$). This simplifies to \begin{equation} 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} . \end{equation} 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}

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Concavity of $\log$ gives $$\log\left( \prod_{i=1}^nx_i^{p_i}\right) = \sum_{i=1}^n p_i\log x_i \stackrel{\mbox{concavity}}{\leq} \log\left( \sum_{i=1}^np_i x_i\right) \stackrel{0\leq p_i\leq 1}{\leq} \log\left( \sum_{i=1}^nx_i\right)$$

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