Successive prime numbers and triangles Except the triplet $(2,3,5)$, I am wondering if there exists another triplet of successive prime numbers such that they do not form a triangle (each prime number corresponds to a side of the triangle).
The question may be asked otherwise:

Are there primes $p_k, p_{k+1}$ and $p_{k+2}$, with $p_k > 2$, such that $p_k + p_{k+1} \leq  p_{k+2}$?

Thanks in advance.
 A: You can prove it in a trivial way if you use the non-trivial result proven by Jitsuro Nagura that for $n\ge25$ there is always a prime between $n$ and $6n/5$. In that case:
$$p_{k+2}<\frac{6p_{k+1}}{5}<\frac{36p_k}{25}\tag{1}$$
On the other side:
$$p_k+p_{k+1}>2p_k\tag{2}$$
By comparing (1) and (2) you get that:
$$p_k+p_{k+1}>p_{k+2}$$
For $n\lt25$ you can do a manual check which is not a problem considering a small number of cases: (2,3,5), (3,5,7), (5,7,11), (7,11,13), (11, 13, 17), (13, 17, 19), (17, 19, 23), (19, 23, 29) and (23, 29, 31).
Obviously (2,3,5) is the only triplet that does not form a triangle.
By the way: the fact that $p_{k+1}<2p_k$ is known as Bertrand's postulate.
EDIT: To prove this inequality you can also use some of the weaker statements proved by Nagura in the same paper:

For example, you can choose the fact that for $n\ge9$ there is always a prime between $n$ and $4n/3$. The procedure stays the same but you'll have to do fewer manual checks.
A: Hint: We have $p_{k+2}<2p_{k}<p_{k+1}+p_k$ for all $k\ge 4$.
Edit: The first inequality follows from a refined version of Bertrand's postulate.
