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I have the optimization problem

$$\begin{array}{ll} \text{maximize} & \displaystyle\sum_{i=1}^n \log(c_i + x_i)\\ \text{subject to} & \displaystyle\sum_{i=1}^n x_i = 1\\ & x_i\ge0\end{array}$$

where $c_i>0$. How could I solve it?

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    $\begingroup$ You should make all of your $c_i$ quite large. $\endgroup$ – T.J. Gaffney Jun 2 '16 at 19:48
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    $\begingroup$ Try using the Lagrange multiplier method. If you do, remember that you have to add n+1 lagrange multipliers(one for each constraint), n to ensure that each x_i is non-negative, and 1 to ensure that they add up to one. $\endgroup$ – Constantinos Vrohidis Jun 2 '16 at 20:17
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Maximizing your function is the same as maximizing its exponential, that is: $$\prod_{i=1}^{n}\left(c_i+x_i\right)\stackrel{AM-GM}{\leq}\left(\frac{1}{n}\sum_{i=1}^{n}(c_i+x_i)\right)^{n} =\frac{1}{n^n}\left(1+\sum_{i=1}^{n}c_i\right)^n.$$ Can you take it from here?

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  • $\begingroup$ Thank you @Jack D'Aurizio! Is it ok if we try to maximize the approximating AM function instead of original, since it is not an equality? Do I need to prove this step somehow? $\endgroup$ – pepelac24 Jun 3 '16 at 20:45

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