It's usual when presenting a theorem to also present its converse. Surprisingly, I've never seen the triangle inequality's converse stated.
Triangle inequality: If the sides of a triangle are a, b, and c, then a + b > c.
The converse of this statement is not obvious to me. I've found two references:
"A type of converse of the triangle inequality holds as well" : http://books.google.com/books?id=EbwNKD0xkUwC&pg=PA12
"A statement often called the converse of the triangle inequality states": http://math.rice.edu/~evanmb/math366spring10/math366hw3.pdf
But they use "a type" and "often called"... And it talks about the existence of a triangle... I'm not sure if it's the converse.
Please explain this to me (I'm very confused).
From the second source you referenced: " A famous theorem of Euclidean geometry, the triangle inequality, states: 'The length of each side of a triangle is less than the sum of the lengths of the other two.' A statement often called the converse of the triangle inequality states: 'Given three lengths a, b and c such that a + b > c, b + c > a and c + a > b, there exists a triangle with side lengths a, b and c.' "
From your statement of the triangle inequality, we may formulate it as the implication $P \implies Q $ where $P$ is the statement: There exists a triangle whose sides are $a$, $b,$ and $c$; $Q$ is the statement: $a + b > c$ (for any choice of $a,b,c$). Then the converse $Q \implies P$ is literally as formulated in the source above. That is, given that $a + b > c$, $b + c > a$, and $c + a > b$ (the statement $Q$), then there exists a triangle with side lengths $a,b$, and $c$ (the statement $P$).