There is a well-known substitution for proving geometric inequalities: If $a,b,c$ are the side lengths of a triangle, then in an inequality involving $a,b,c$ it is possible to replace $a,b,c$ by $x+y,y+z,z+x$. This gets rid of the "triangle" restriction, so you just have to prove the resulting inequality in $x,y,z$ for any positive reals $x,y,z$.
My question concerns the other direction: what if you have an inequality in $x+y,y+z,z+x$, is it possible to make the substitutions $x+y=a$, $y+z=b$ and $z+x=c$ and end up with an inequality in $a,b,c$ that is valid for the side lengths of a triangle?
Yes.
If you have an inequality in positive reals $x, y, z$, then by defining $a=x+y, b= y+z, c = z+x$, we get positive reals $a, b, c$ s.t. $a+b - c = 2y > 0, b+c-a=2z> 0, c+a-b=2x > 0$, so $a, b, c$ form a triangle.