Proving a relation with induction I have a problem:
Let $p_n$ be the $n:th$ prime number ($p_1=2,  p_2=3,  p_3=5$ and so on). With induction, show that $p_{n+2}>3n$ for each integer $n\geq1$.
I can't figure this out because the primes are confusing me, making me unable to show the inductive step.

For $n=1$ we can verify that $p_3=5>3\cdot1$
Suppose that the statement holds true for $n$ meaning that $p_{n+2}>3n$. We will show that
this holds for $n+1$.
 We have $p_{(n+1)+2}=p_{n+3}$. We already know from the induction step that $p_{n+2}>3n$  which means $p_{n+2}\geq 3n+1$.
So if $p_{n+3}\leq 3(n+1)=3n+3$ we have only one  possibility: $p_{n+3}=3n+2$   

(or $p_{n+3}=3n+1$ which means that $p_{n+2}$ and $p_{n+3}$ coincide .
  Also, if $p>3$ then $p$ cannot be  a multiple of $3$ so the case
  $p_{n+3}=3(n+1)$ is rejected)

But this means that $p_{n+2}$ and $p_{n+3}$ are consecutive primes which is only possible if the primes are $2,3$ (we are already above $5$)
All at the above step arrive at a contradiction and so we must have only one possibility: $p_{n+3}>3(n+1)$   
this proves the case for $n+1$ and so, for every $n$.
