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

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


$$S\leq b_1 x_1$$

$$S\leq b_2 x_2$$


$$S\leq b_n x_n$$

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

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.