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I have the following equation.

$$x_1+x_2+x_3+x_4+x_5=1$$

Where $x_1$ is the largest, $x_2$ is the second largest and so on. We have ten triplet combinations of these values which we can create. We apply the triangle inequality to the triplets. For example,

$$x_1>x_2+x_3$$

$$x_2>x_3+x_4$$...etc.

What is the minimum amount of the ten inequalities which can be satisfied. Is it possible that all ten of the inequalities cannot be satisfied?

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  • $\begingroup$ If your question is based on another one, please put the link back to the old question. In many cases, this will help answerers understand more about the question and what has been done. $\endgroup$ – Toby Mak Nov 21 '17 at 0:15
  • $\begingroup$ Your triangle inequalities have the wrong directions. For example, for $x_1,x_2,x_3$, the triangle inequality is $x_1 < x_2 + x_3$. $\endgroup$ – quasi Nov 21 '17 at 0:16
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Yes, it's possible for no triangle inequalities to be satisfied.

For example, let $$(x_1,x_2,x_3,x_4,x_5) = (16t,8t,4t,2t,t)$$ with $t$ positive, but not yet specified.

It's easily seen that for those $5$ values, no $3$ distinct values can be used as lengths to form a triangle.

Now just solve the equation $x_1 + x_2 + x_3 + x_4 + x_5 = 1\;$for $t$.

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Example : \begin{eqnarray*} x_1= \frac{17}{38},x_2= \frac{10}{38},x_3= \frac{6}{38},x_4= \frac{3}{38},x_5= \frac{2}{38}. \end{eqnarray*} One can rapidly verify that $x_i> x_j+x_k$ for $i<j<k$.

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