Capturing the three aspects of the Triangle Inequality in a single condition 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)?     
 A: 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. 
A: 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 $x + y > z$.
