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.

Solution 1:

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.

Solution 2:

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..