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Suppose in an optimization problem we have the following constraints $$\sum_{i=1}^K\frac{1}{x_{i}}\leq T,$$ $$0<x_i<X$$ where $X,T$ are some real constants. In this case, is it ok to use the following relation (due to Jensen inequality) $$K\frac{K}{\sum_{i=1}^Kx_i}\leq \sum_{i=1}^K\frac{1}{x_i}$$ and use the fact that the above Jenson inequality becomes equal when all $x_i$'s are equal. Then, replace the original constraint $\sum_{i=1}^K\frac{1}{x_{i}}\leq T$ in the optimization problem with the new constraint $$K\frac{K}{\sum_{i=1}^Kx_i}\leq T$$ with obviously all $x_i$'s in this constraint being equal. How we can justify this trick? Any help in this regard will be much appreciated. Thanks in advance.

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There is no trick, if $a\leq b$ and $b\leq c$ then $a\leq c$:

$$K\frac{K}{\sum_{i=1}^Kx_i}\leq \sum_{i=1}^K\frac{1}{x_i}\;\;\;\;{\rm and }\;\;\;\;\;\sum_{i=1}^K\frac{1}{x_{i}}\leq T,$$ so $$K\frac{K}{\sum_{i=1}^Kx_i}\leq T$$

But of course, this is now weaker constrain then starting one.

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  • $\begingroup$ thank you for your comment. Since you have mentioned that its a weaker constraint then it means that we can not say that if we replace the original constraint with the new constraint then the new optimization problem is equivalent to the original one. Right? $\endgroup$ – Frank Moses Jan 29 '19 at 9:14
  • $\begingroup$ Yes, that is correct. $\endgroup$ – Aqua Jan 29 '19 at 9:28
  • $\begingroup$ But is there a condition in which the two problems will be equivalent. (For example if we prefer to have smaller values of $x_i$'s etc)? $\endgroup$ – Frank Moses Jan 29 '19 at 10:12

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