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Let $$ \pi = f_1(x_1) + f_2(x_2) + f_3(x_3) + \dots + f_n(x_n) = \sum_{i=1}^n f_n(x_i) $$ where $f_i$ denote different functions and $x_i$ denote different independent variables

Would proving that $\pi$ is maximized by maximizing $f_i(x_i) \forall i$ be as simple as assuming, BWOC, that $\pi$ is not maximized by doing so, and then noting that this is a contradiction to $f_i(x_i)$ being maximized for each term, since some term must be higher?

Note that if this was written as an optimization problem, it would be unconstrained (as I believe constraints would mess with this result).

It is pretty much a trivial question/property, but I see it used here and there and have not seen why it is true discussed.

Edit: Additionally, what if each term was instead the composition of functions? That is, what if we had $$ \pi = g(f_1(x_1)) + g(f_2(x_2)) + g(f_3(x_3)) + \dots + g(f_n(x_n)) = \sum_{i=1}^n g(f_n(x_i)) $$ I believe the same argument applies, correct? Because, once again, if maximizing each term does not maximize the sum, then some term must be larger than the maximum, which is a contradiction.

Edit 2: and it would also then apply to multivariable functions, so long as each function is of different (multiple) independent variables.

Thanks.

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Hint:

$$\frac{\partial \pi}{\partial f_i} = 1 > 0$$

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  • $\begingroup$ While this is a good and useful insight, it seems likely that the OP requires a more detailed explanation of how the fact leads to the desired conclusion. $\endgroup$ – hardmath Feb 1 '16 at 2:38
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    $\begingroup$ @hardmath, no , it is fine. I realized this an hour or so after posting the question but did not get around to answering the question myself. $\endgroup$ – majmun Feb 1 '16 at 16:03
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    $\begingroup$ @majmun: In that case I will up vote on your behalf! $\endgroup$ – hardmath Feb 1 '16 at 16:15

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