Prove that every integer $n\geq 7$ can be expressed as a sum of distinct primes. My teacher said to use Bertrand's postulate and I have tried this for so long and I seem to go nowhere. Help would be appreciated.
EDIT: Here's what I've done in my proof so far (I need help finalizing case 2)
First note that for $p_0=13$, we can express all integers $7 \leq n \leq 2p_0=26$ as a sum of distinct primes less than or equal to $13$. 
Now, we will prove that we can construct these sums indefinitely. Assume we know some prime $p'$ exists such that every integer $7\leq n \leq 2p'$ can be expressed as a sum of distinct primes less than or equal to $p'$. Then, by Bertrand's postulate, there exists a prime $p$ such that $p'<p<2p'$. We claim now that every integer $7\leq n \leq 2p$ can be written as a sum of distinct primes less than or equal to $p$. Consider the two following cases
Case 1: $2p'-p\geq 7$, hence $2p'\geq p+7$ so the terms $p,p+1,\dots,p+7$ are less than or equal to $2p'$ which means they can be written as a sum of distinct primes $\leq p'$ by hypothesis. It is left to check whether the terms $p+8,p+9,\dots, 2p$ satisfy our claim. Note if we subtract $p$ from every term in the arithmetic progression above, it becomes $8,9,\dots, p<2p'$ which shows that each term can be written as the sum of $p$ plus some other distinct primes less than or equal to $p'<p$ by hypothesis.
Case 2: $2p'-p\leq 6$, hence $2p'\leq p+6$. Here we can see all terms $p+7,\dots, 2p$ satisfy our claim along with $p+2,p+3$ and $p+5$ by a similar argument as in Case 1. I'm not sure how to deal with $p+1,p+4,$ and $p+6$ though.
EDIT: Oh, since any prime $p \geq 13$ is odd, then the only possible values for $2p'$ in Case 2 are $p+1,p+3$ and $p+5$, so I don't have to worry about the other cases. I think I'm done!
EDIT: Nope, I still need to deal with $p+4$ and $p+6$.
 A: We shall inductively prove a stronger form, namely that every positive integer $n \ge 7$ can be written as the sum of distinct primes such that the largest is at most $\max(11,n-7)$. It turns out that strengthening makes the induction work!
Take $n \ge 28$.
Let $m = \lceil \frac{n-6}{2} \rceil= \lfloor\frac{n-5}{2} \rfloor$.
Let $p$ be a prime such that $m+1 \le p \le 2m-1$ [by Bertrand's postulate].
Then $\frac{n-5}{2} \le p \le n-7$.
By the induction to be established $n-p$ can be written as a sum of distinct primes such that the largest is at most $\max(11,n-p-7)$, which is less than $p$ because $p \ge \frac{28-5}{2} > 11$ and $2p \ge n-5 > n-7$.
Thus $n$ can be written as a sum of distinct primes such that the largest is at most $\max(7,n-7)$, and the induction holds as long as the claim is true for every $n$ from $7$ to $27$.
7 = 5+2
8 = 5+3
9 = 7+2
10 = 5+3+2
11 = 11
12 = 7+5
13 = 11+2
14 = 7+5+2
15 = 7+5+3
16 = 11+5
17 = 7+5+3+2
18 = 11+7
19 = 11+5+3
20 = 11+7+2
21 = 11+7+3
22 = 13+7+2
23 = 13+7+3
24 = 13+11
25 = 13+7+5
26 = 13+11+2
27 = 13+11+3

A: Hint: Bertrand's postulate is useful for the inductive step.  i.e. with the assumption this holds for all $n < M-1$, you have some prime $p$ s.t. $M \ge p > M/2$.  Thus you can express $M-p$ as a sum of distinct primes, none of which are $p$.   
All that remains is to show $n=7$ and that cases for which $M-p < 7$ can be handled. $M-p \in \{1, 4, 6\}$ will merit some detail... (times like this one wishes $1$ is allowed as a prime.  This last part however seems to be quite difficult).
