good one.

Consider a permutation P_\propto (\lambda) of the set \lambda = \left(\alpha_1, \alpha_2....\alpha_7 \mid 2\le \alpha_i \le 12 : 1\le i\le 7 \right)

P_\propto (\lambda)=(a_1,a_2,a_3...a_7 \mid a_i\le a_{i+1})

Now think of the triplets - \left\{\begin{matrix} (a_1,a_2,a_3) & (a_2,a_3,a_4) & (a_3,a_4,a_5) & (a_4,a_5,a_6) & (a_5,a_6,a_7) \end{matrix}\}\right.

Needless to mention that (in the ith triplet) a_i+1+a_i+2>a_i & a_i+a_i+2>a_i+1....The only thing to prove remains that there exists an i such that a_i+a_i+1>a_i+2.....

The proof is obviously by contradiction -

By assuming the opposite, we have \sum_{i=1}^{5}\left({a_i+a_{i+1}}<a_{i+2} \right)

Simplyfying gives \sum_{i=1}^{5}{a_i}<a_1+2a_2+\sum_{i=3}^{5}{a_i}<a_7

Which is only possible if a₇=12 and a₁=a₂=2....(Assuming no 3 are equal)....which implies \sum_{i=3}^{5}{a_i}<8 - An obvious contradiction....

Edit : Assuming all the nos. are integers....(Which is not specified....h4hemang is requested to confirm it)

Why assume what is not said [7]

Of course he meant reals and such a proof is easy to construct by PHP. take [2,4],(4,8],(8,12]. One box must contain 3 numbers and they form sides of a triangle...

perhaps because its very weak...

one can easily extend (2,12) to (2,16)

i don't know the largest k , so that its valid for (2,k) though...

seems interesting

the result is valid for (1,8), and one can extend it to (1,13) ,again i dnt know the upper bound

Oh well...

fix any positive real n. if you choose k element from the set [n,t), so that at least 3 among the chosen elements form sides of a triangle then maximum value of t is given by n(F_k) where F_k is the k-th fibonacci number. For t > n(F_k) there will be k numbers ,so that no three of them forms sides of a triangle... for example n,n,2n,3n,5n,...n(F_k).so its the strongest bound.

@Devil: why do you want to shrink the upper bound? of course the result is valid for all n<t≤n(F_n)...didn't get what you meant.

If you mean the minimum no of numbers to be chosen form (2,12) for ensuring existence of triangle ,then it would be 6. (so no need to choose 7 numbers). actually choosing 6 numbers from (2,16) is the strongest case...

so h4hemang can make is question strongest by restating it:

If any 7 numbers( not necessarily distinct ) are chosen from 2 to 26, prove that among those 7 numbers we can get three which form the sides of a triangle.