Seven line segments, with lengths no greater than 10 inches, and no shorter than 1 inch, are given. Seven line segments, with lengths no greater than 10 inches, and no shorter than 1 inch, are given. Show that one can choose three of them to represent the sides of a triangle.
 A: Hint given by @MJD was really helpful
If you number the line segments from 1 to 7, and suppose no three of them can be used to make a triangle.
Start from seg. 1 and seg.2 . As you cannot use seg. 3 to make a triangle from seg.1 and seg. 2 so length of seg. 3 must be strictly more than 2 inches.
Continue same way you shall get (strict) lower bounds on lengths for 
seg. 1 ------ 1 inch
seg 2 --------1 inch
seg 3---------2 inch
seg 4---------3 inch
seg 5---------5 inch
seg 6---------8 inch
Seventh segment cannot be greater than 10 inches, then it can be used to make a triangle with two of the previous 6 segments. (seg. 5 and seg. 6 for example)
Now question how to use Pigeonhole principle in here...?
I am sorry I did not answer your question completely
A: Let the lengths of seven of the line segments be $l_1, l_2, l_3, \dotsc, l_7$ in (weakly) ascending order.
Now suppose that no triangle can be formed from these lengths.
Furthermore, let $l_2 ≥l_1 ≥1$ (the minimum length,suggested by meelo).
Then $l_3 > 2$, $l_4 > 3$, $l_5 > 5$, $l_6 > 8$, $l_7 > 13$ (because if $l_i + l_{i+1} \geq l_{i+2}$, then we can form a triangle).
But we know that $l_i \leq 10$, so thus a triangle has to be formed.
A: Hint: Suppose $a < b < c$ are three lengths that do not form a triangle.  Then $c > a+b > 2a$.
A: Minimum sequence (degenerate triangles):
1, 1, 2, 3, 5, 8, 13
If any number were reduced a triangle would form. Therefore to make no triangles with 7, the upper bound must be 13.
I don't see how the pigeonhole principle even applies.
