Do 3 consecutive primes always form a triangle? Suppose that $a$, $b$, and $c$ are any three consecutive primes other than the triple $2$, $3$, and $5$.  Do they satisfy the triangle inequalities: $a + b > c$;  $b + c > a$;  $c + a > b$?  In other words, can we always form a triangle with sides being the $3$ successive prime numbers?  Is this a well-known result?  Where can I read about its proof or refutation?  Thanks in advance.
 A: Hint: Use a stronger form of Bertrand's postulate, which states that $p_ {n+1} < 1.1 \times  p_{n}$ for large enough $n$.
As such, $p_{n-1} + p_{n} > p_{n+1}$ satisfies the triangle inequality.
This means that we only need to check finitely many small cases, which is easy to do.
A: Let $p_n,p_{n+1}, p_{n+2}$ be three consecutive primes.
You need to show that 
$$p_{n+2}< p_{n}+p_{n+1}$$
This follows immediately from the following stronger version of the Betrand Postulate: For $n \geq 7$ there are two primes between $n$ and $2n$.
The case $2,3,5$ is obviously a counterexample, an easy check up to 7 shows there is no other.
P.S. Does anyone have a good reference to this stronger version of the BP?
It follows from last statement of this Paper, but I remember seeing once that statement written explicitely.. 
A: I just love answering old questions :) ... basically:


*

*Bertrand's Postulate $p_{n+1} < 2 \cdot p_n$

*From (Limit inferior of the quotient of two consecutive primes) $$\lim_{n \to \infty } \frac{p_{n+1}}{p_{n}}=1 \Rightarrow \lim_{n \to \infty } \frac{p_{n+1}}{p_{n-1}}=\lim_{n \to \infty } \frac{p_{n+1}}{p_{n}} \cdot \frac{p_{n}}{p_{n-1}}=1$$  or $$p_{n+1} < 2 \cdot p_{n-1}$$ from some $n$.


As a result:
$$p_{n-1}+p_{n}\geq 2 \cdot \sqrt{p_{n-1} \cdot p_{n}}= \sqrt{2 \cdot p_{n-1} \cdot 2 \cdot p_{n}} > p_{n+1}$$
