Check me on this; I think it's pretty straightforward (maybe someone already posted this proof):

Show 14 is the least m such that for all n >= m, there are j and k such that n = 3j+8k.

First, show there is no m < 14 such that for all n >= m, there are j and k such that n = 3j+8k.

Try every m less than 14, finding the least n >= m such that there are no j and k such that n = 3j+8k. To prove, for each m < 14 and for its selected n, that there are no j and k such that n = 3j+8k, just try all j and k less than n (since j and k would have to be less than n in order for n = 3j+8k to hold).

Show for all n >= 14, there are j and k such that n = 3j+8k.

First we show it for n is even.

By weak induction on r, it's easy to show: If r > 3 then there is a q such that 2r = 2q+8.

Then, by strong induction on r, show: If r > 7, then there are j and k such that 2r = 3j+8k, as follows:

For the stong induction hypothesis, suppose for all q such that 7 < q < r, we have there are j and k such that 2q = 3j+8k.

Now for some q, let 2r = 2q + 8. So q < r. And, since r > 7, we have q > 2. Now if q is in {3, 4, 5, 6, 7} then it's easy to verify that, for each case, there are j and k such that 2r = 3j+8k. So suppose 7 < q. So let 2q = 3j+8k. So 2r = 3j+8(k+1).

So, for n is even, we're done.

Now, suppose n is odd. Checking n = 15 and n = 17 is easy. So, suppose n > 17.

By weak induction on n, it is easy to show that for every n such that n is odd and n > 17, there is an r such that n = 2r+3.

So for some j and k, we have n = 2r+3 = 3(j+1)+8K.