Question 574686
The proof is a bit weak, because there is no base case, and I don't really see how they obtained 2k + 1 > 2 + 2k (multiplying by 2, you get 4k > 2+2k. You can say 4k = 2k+2k > 2k+(1+k) > 2k+1, but you can't claim that 2k+1 > 2+2k [which *cannot* be true]).


An induction proof would assert that the base case, n = 3, is true (which it is). If the inequality holds for some k >= 3, then we can show it holds for k+1.


*[tex \LARGE 2(k+1) > 1 + (k+1) \Leftrightarrow 2k + 2 > (k+1) + 1 \Leftrightarrow 2k - (k+1) + 1 > 0]


However, we already know that 2k - (k+1) > 0 because 2k > k+1 (this is assumed to be true for our induction) so the above statement must be true. ∎


Of course, the simplest way to prove it is to say that


*[tex \LARGE 2n > 1+n] if and only if *[tex \LARGE n > 1] (subtracting both sides by n). However this is definitely true since n > 2, so we're done. ∎