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I would like to minimize the following function:

$f(x)=log(e^{-x_1}+..+e^{-x_n})$

Subject to:

$\sum_{i=1}^{n}{x_i}=1$

$0 \leq x_i \leq 1$

So far I have discovered the following: If all the $x_i$'s are equal, $f(x)=max(x_i)+log(n)$, but I have not been able to find a conclusive answer regarding what would be the value of the $x_i$'s for the minimum value of the function.

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  • $\begingroup$ Min occur at $x_i = \frac{1}{n}$. $\endgroup$ – Mhenni Benghorbal May 4 '14 at 20:58
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Let $y_i = e^{-x_i}$, then $0 < y_i \leq 1$, and $x_i = -lny_i$, and then $-lny_1 - lny_2 -...-lny_n = 1 \to ln(y_1y_2....y_n) = -1 \to y_1y_2...y_n = \dfrac{1}{e}$. So: by AM-GM inequality: $log(y_1 + y_2 +...+ y_n) \geq log(n\cdot (y_1y_2...y_n)^{\frac{1}{n}}) = log(n\cdot e^{-1/n}) = logn - \dfrac{1}{n}$.

So: $f_{min} = logn - \dfrac{1}{n}$

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  • $\begingroup$ Is this a lower bound on the minimum? Or is this the minimum? From the proof it seems to be a lower bound on the minimum. $\endgroup$ – ASDF May 5 '14 at 16:59
  • $\begingroup$ @ASDF: it is the minimum. $\endgroup$ – DeepSea May 5 '14 at 19:33
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log is increasing, so the minimum occurs at the same place as the minimum of $\sum e^{-x_i}$; and $x\mapsto e^{-x}$ is convex, so $\sum e^{-x_i} \ge ne^{-\frac1n\sum x_i} = ne^{-1/n}$, with equality when all $x_i$ are equal.

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