Lengths of the sides of a triangle: sufficient and necessary condition? For any three positive scales, $a,b,c$, what is the sufficient and necessary condition such that they can form a triangle?
Is $a+c>b,a+b>c,b+c>a$ enough?
Thanks!
 A: Assertion:  The system of inequality is both necessary and sufficient, for the existence of a triangle with side lengths = a, b and c.
Proof:
-- Necessity
The shortest path between any two points is the straight line connecting such two points.
-- Sufficiency
Rank the three numbers.  Without loss of generality, let a be the number that is no smaller than the other two, i.e. a >= b and a >= c.
At any point O, draw two concentric circles with radius a and b respectively.
(OK for them to coincide in the case of a = b.)
Draw a horizontal diameter intersecting the outer circle at Al(eft) and Ar(ight), inner circle at Bl and Br. 
Then distance AlBl = a-b and AlBr = a+b.  Since a-b < c < a+b, then, by continuity, there must exist a point on the inner circle (denoted as B) such that AlB = c.  
Further, points Al, O and B cannot possibly be co-linear (on a straight line) as that would imply that a = b + c.
Al-O-B thus forms the desired triangle.    
Q.E.D.
A: Condition
$a=x+y,\ b=y+z,\ c=z+x\land x,y,z\in \mathbb{R^+}$
Necessity
$a=x+y,\ b=y+z,\ c=z+x$
$a+b>c \implies x+2y+z>z+x \implies y>0$
$b+c>a \implies z>0$
$c+a>b \implies x>0$
Sufficiency
$y>0 \implies x+2y+z>z+x \implies a+b>c$
$z>0 \implies y+2z+x>x+y \implies b+c>a$
$x>0\implies z+2x+y>y+z\implies c+a>b$
