Proof by induction that $n \ge 3, n^2 \ge 2n+1$

Question

I wrote a basic proof by induction but did not use the inductive hypothesis. Not sure if this proof is correct.

1. Base case: $3^2 \ge 2\times3 + 1 $
2. By inductive hypothesis, $k^2 \ge 2k+1 \space\forall k\ge 3$
3. Inductive step: $(k+1)^2 \ge 2(k+1)+1 \rightarrow k^2+2k+1 \ge 2k+3 \rightarrow k^2\ge2$ is clearly true for all $k \ge 3$

I think not using the inductive hypothesis makes me uncomfortable about this proof's correctness. I don't know how to check if there's circular logic here.

Answer 1

On some manipulation, the Inequality becomes $$ n^2 -2n + 1 \geqslant 2 \quad [\text{for } n \geqslant 3] $$ Which is obviously $(n -1)^2 \geqslant 2$ for $n \geqslant 3. $ Which is always true for $n \geqslant 3$.

Answer 2

Your ans is correct. All you need to show in induction is$~$ "If $p_n$ is True, then $p_{n+1}$ is True".

If $p_{n+1}$ is always TRUE (which you proved) then you are done as above assertion will be true then by default.

Please clearly note the logical relation between If and Then - to understand this. (So, in other words, if you have proved something for a universal set - and you are asked If that will be true for a subset - Ans will be yes.)