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We are given positive values $b_1, b_2,\dots,b_n$ and we need to find if there exist $n$ positive numbers $x_1, x_2, x_3,\dots,x_n$ such that if

$$x_1+x_2+\cdots+x_n =: S$$

then

$$S\leq b_1 x_1$$

$$S\leq b_2 x_2$$

$$\vdots$$

$$S\leq b_n x_n$$

Is there mathematical solution to this problem or some direct formula? We just need to state if sum can satisfy all inequalities or not .

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  • $\begingroup$ Are there any restrictions on the $b_i$ values? Can they be positive or negative? or just positive? Check out this for help with mathjax formatting. Good formatting makes your question much more readable (and therefore immediately a better question). $\endgroup$ – TravisJ Aug 18 '16 at 12:31
  • $\begingroup$ They can only be positive . $\endgroup$ – satyajeet jha Aug 18 '16 at 12:35
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    $\begingroup$ I've made an edit to your post. Please take a minute and look at what I did with the formatting, it will help you in the future. $\endgroup$ – TravisJ Aug 18 '16 at 12:37
  • $\begingroup$ It depends on the given numbers $b_i$. If $b_1=1$ and $n>1$ then it is impossible. If $b_1=\dots =b_n=n$ the one can take $x_1=\dots =x_n=1$. $\endgroup$ – Robert Z Aug 18 '16 at 12:45
  • $\begingroup$ We need to find a general solution to this problem . $\endgroup$ – satyajeet jha Aug 18 '16 at 12:47
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Clearly, all $b_i>0$.

Suppose that $\{x_i\}$ is a solution. Then it is easy to see that $\{a x_i\}$ is also a solution for any positive $a$ since the equality and all the inequalities are linear. Thus, without any loss of generality, we can set $S=1$.

Then each of the inequalities reduces to $x_i \ge 1/b_i$. Combining the inequalities with the equality, we obtain the necessary condition$$\sum\limits_i 1/b_i \le 1$$

This condition is also sufficient. If this is satisfied, then there is always a solution. One possible solution is $$x_i = \frac{1/b_i}{\sum\limits_j(1/b_j)}$$

For this solution, it is easy to see that $$S=\sum\limits_i x_i=1$$ and $$b_i x_i = \frac{1}{\sum\limits_j(1/b_j)} \ge 1 = S$$ Thus all the constraints are satisfied.

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