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So in a triangle, where the sides are length $x$, $y$, $z$, the three aspects of the Triangle Inequality, (1) $x+y>z$, (2) $x+z>y$, and (3) $y+z>x$, must be satisfied in order for the triangle to be valid.

If we sum up all the conditions, we get $2(x+y+z)>x+y+z$; that is, $2>1$. Which doesn’t tell us much. Is there a condition/formula that one could use that checks all three conditions to tell us if the sides make a valid triangle? Any way to combine statements (1)+(2)+(3)?

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    $\begingroup$ jessica: Please don't change the initial question asked. If you want to add information or clarify, do so after the initial question, writing EDIT <edit following> or such $\endgroup$
    – amWhy
    Sep 24, 2013 at 20:09
  • $\begingroup$ The question now seems to be: starting from the numbers $x,y,z$, is there a triangle with those sides? (Rather than that we already have a triangle, and etc... But if we already have a triangle then all three inequalities hold definitely, so it seems there is no question.) $\endgroup$
    – coffeemath
    Oct 10, 2013 at 11:04

2 Answers 2

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Given the positive numbers $x,y,z$, define the function $T(x,y,z)$ by $$T(x,y,z)=(x+y-z)(y+z-x)(z+x-y).$$ Then if $T=0$ the only possible triangle formed by sides $x,y,z$ is degenerate (i.e. the three vertices must lie on a line and it's not really a triangle. If $T>0$ then there is a nondegenerate triangle with sides $x,y,z$, and if $T<0$ there is no triangle with sides $x,y,z.$

This may be shown by first considering that if the sides are arranged as $x \le y \le z$ then a nondegenerate triangle is formed iff $x+y>z$ while a degenerate one is formed if $x+y=z$ and no triangle results if $x+y<z$.

Now if $x\le y \le z$ and $x+y>z$, then automatically also $x+z>y$ and $y+z>x$, so that $T>0$. Also given $x \le y \le z$, even with no more assumptions we automatically have (in the nondegenerate case) $y+z >x$, and if we throw in also the assumption that $x+y<z$, we have one negative factor $x+y-z$. So to finish we need to show the other factor $x+z-y$ is positive under these assumptions, so that $T$ comes out negative when there is no triangle. But if both $x+y<z$ and $x+z<y$, we get the contradiction that $y<z-x$ and $y>z+x$ at the same time, impossible since $z-x<z+x.$

In summary, there is a single formula to check the numbers $x,y,z$ to see if they can be sides of a triangle. Of course in applying this formula one would already be calculating each of the inequalities separately anyway, so it's arguably only a curiosity.

Added: A shorter proof. Note first that $T$ is invariant under the set of all six rearrangements of $x,y,z$ [this is easy to see, or one can just multiply $T$ out completely then it's very clear]. So there is no loss on assuming that $x \le y \le z.$ Name the first factor $A=x+y-z$ and the other two $B=y+z-x,\ C=z+x-y.$

From $x \le z$ we have $z-x \ge 0$ so that $B=y+(z-x)\ge y>0.$ Similarly from $y \le z$ follows $z-y \ge 0$ so that $C=x+(z-y)\ge x>0.$

Now $T$ is the product $ABC$ and we've shown from $0<x \le y \le z$ that each of $B,C$ is definitely positive. It follows that everything is determined by factor $A,$ i.e. that there is a nondegenerate triangle if and only if $T>0$ as claimed.

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  • $\begingroup$ 1) Is there a name for this function T i.e. who discovered, and when this function was discovered? 2) the assertion "if the sides are arranged as …sides x, y, z" seems intuitive, but is there a way to mathematically prove this. Thank you. $\endgroup$ Oct 6, 2016 at 12:23
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    $\begingroup$ @Globalnomad If one checks all the rearrangements of $x,y,z,$ the value of $T(x,y,z)$ is the same for each rearrangement. So the assumption above that $x \le y \le z$ for the sides of the triangle doesn't restrict the validity of the argument. Oh, and no I don't know if there's a name used for this function $T,$ though it wouldn't surprise me since it is somewhat symmetrical. Apart from one other factor of $x+y+z$ it appears under the radical in an adjusted form of Heron's formula for the square of the area of a triangle. $\endgroup$
    – coffeemath
    Oct 6, 2016 at 23:10
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    $\begingroup$ @Globalnomad Note that I have now added a much shorter (and hopefully clearer) proof that $T>0$ characterizes nondegenerate triangles for positive sidelengths $x,y,z.$ $\endgroup$
    – coffeemath
    Oct 7, 2016 at 5:11
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Any one of the conditions suffices to prove that we have a non-degenerate triangle. That is, three points with respective lengths between points given by $x, y, z$ (we are free to permute the labels of the side-lengths) are non-colinear (form a non-degenerate triangle) if and only if $x + y > z$.

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  • $\begingroup$ Right, but is there a condition/formula that one could use that meets all three conditions to tell us if the sides meet the condition of a triangle? $\endgroup$
    – jessica
    Sep 24, 2013 at 19:08
  • $\begingroup$ Any one of the conditions being met $⟹$ the other two conditions are met. I.e., one condition being met $⟺$ the lengths of the sides meet the conditions of being a triangle. $\endgroup$
    – amWhy
    Sep 24, 2013 at 19:25
  • $\begingroup$ How about 7,16,8. 8<7+16. But 7+8<16. So obviously checking one isn’t sufficient $\endgroup$
    – jessica
    Sep 24, 2013 at 19:56
  • $\begingroup$ You posted that we have a triangle! The distances between any three non-colinear points will always be such that conditions 1, 2, 3 are all met. So if we know any one condition is met, we know that all three are met. At any rate, the placement of the points determine the lengths of the sides. $\endgroup$
    – amWhy
    Sep 24, 2013 at 19:59
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    $\begingroup$ This answer is wrong or I cannot understand it. There is no triangle with side lengths $x=1$, $y=100$, $z=2$, but in such a case $x+y>z$. If you had said something like, reorder in increasing order of size, $x\leq y\leq z$, then it would make sense, although it could be added how the reordering by size is done. $\endgroup$ Oct 13, 2013 at 3:10

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