Proof of Existence of a (Planar) Triangle Given Three Sides Suppose that a triangle is required to have perimeter $1$, with sides $a,b,c.$ I would like to show the following:
$$a+b\ge .5$$
$$b+c\ge .5$$
$$c+a\ge .5$$
I have already proven necessity, but I don't immediately see sufficiency.
To prove necessity, assume that $a+b < .5$ but then $c\ge .5$, now draw a "degenerate triangle" where the three segments all fall on a single line, this cannot produced a triangle (the image is not "closed", there are "three" vertice). And of course, in the nondegenerate case the triangle still can't be produced.
I was thinking of using law of cosines somehow to prove sufficiency, but not sure..... thanks!
 A: You won't find sufficiency unless one or two of the inequalities is strict:
E.g. $$a+c = 1,\quad a+b = 0.5,\quad b + c = 0.5\;$$ satisfies the inequalities you've written, but they are necessarily all points on a line: on a generate triangle of "length/perimeter" $= 1.$  
A: Assuming you don't care about the marginal degenerate case amWhy refers to - I prefer a constructive geometrical approach to proving existence.
First of all, start by drawing a line of length $a$ horizontally. Now you attach (WLOG) $b, c$ to the left, right of $a$. Where are the possible endpoints for these lines? They form a nice pair of loci. Where these intersect must be where you can put the final vertex.
Then you just need to use your relations to check whether or not they do, in fact, intersect! You will need all three relations (unless you're clever about which edge you choose first).
A: As amWhy pointed out, you might need strict inequalities. Assuming they are strict,
Asume $0 \lt a \le b \le c$ and $a+b+c = 1$
Now $a + b \gt 0.5 \implies a + b \ge \frac{a+b+c}{2} \implies a+b \gt c$
Thus $a,b,c$ are sides of a triangle.
If you allow degenerate triangles, then you can drop the strictness.
