Proof by induction. Two different ways. Which one is valid? Ex. 1
Let's say we want to prove that $n^2 > 2n, \qquad \forall n \geq 3 \in \mathbb{N}$. So first, we would check our base case $n = 3$. Obviously we have $3^2 > 2\cdot 3 = 9 > 6$.
Now our inductive step. We have $\ k^2 > 2k\ $. We add $\ 2k+1 \ $ on both sides, yielding $\ k^2 + 2k + 1 > 2k + 2k + 1$. Because we have $2k > 1$, we can add a $1$ on the right side of the inequation. We have $k^2 + 2k + 1 > 4k + 2 = (k+1)^2 > 2(k+1)$ $\blacksquare$
Ex. 2
Now if we expand $(n+1)^2 > 2(n+1)$ we have $n^2 + 2n + 1 > 2n + 2$. Since we assume $n^2 > 2n$ to be true, could we simply subtract this on both sides? This would leave us with $2n + 1 > 2$, which is trivially true for any $n \geq 2$. Is this valid, and is this enough to prove the original statement?
 A: Ex. 1: After getting $x^2+2k+1>2k+2k+1$, you cannot simply add $1$ to the right side (unless you add it to the left side too). Besides, $4k+2\ne2(k+1)$. But, since $2k+1$, $2k+2k+1>2k+2=2(k+1)$.
Ex. 2: I don't know why you are assuming that $n^2>2n$, but if you are… so what? You have $4>3$ and $5>1$, but you don't have $4-3>3-1$.
A: In your first proof, you should not write things like

$3^2>2\cdot 3=9>6$

because $2\cdot 3\neq 9$. You probably mean $3^2>2\cdot 3\Leftrightarrow 9>6$, showing that the inequalities are equivalent. There is also a small mistake at the end; you cannot just add $1$ to the right side, but you can replace the summand $2k$ by $1$ making the sum smaller, and thus getting $$(k+1)^2=k^2+2k+1>2k+2k+1>2k+1+1=2(k+1)$$
$\ $
Your second proof is kind of correct, but badly written. You should never start with the statement you want to prove and transform it to something that is true, since a wrong statement can imply a true one as well. You could basically write your prove backwards: Since $2n+1>2$, we get $\ldots$
Alternatively, another save way is to start at one side and manipulate it until you are done. For example you could write $$(n+1)^2=n^2+2n+1\overset{\substack{\text{Inductive}\\\text{hypothesis}}}>2n+2n+1\overset{2n+1>2}>2n+2=2(n+1)$$ which basically mimics your proof, but is written in a neater way.
A: Ex. 1

*

*You wrote "$n^2 > 2n, \qquad \forall n \geq 3 \in \mathbb{N}$". This is problematic for two reasons: if you use the symbol $\forall$ rather than the words "for all", then this symbol must be placed before the formula in question; also, $\forall n\ge3\in\Bbb{N}$ reads as "for all $n$ greater than $3$, which is a natural number". You want to say that $n$ is a natural number, not $3$, and so should rearrange things. Both of the following are correct (though I personally prefer the second):
\begin{align}
\require{cancel}
&\forall n(n\in\Bbb{N} \text{ and }n\ge3\implies n^2>2n) \\
&\forall n\in\Bbb{N}(n\ge3 \implies n^2>2n) \, .
\end{align}
Even better is to use words: "every natural number greater than or equal to $3$ satisfies $n^2>3n$".

*For the base case, presumably you meant to write $3^2=9>2 \cdot 3=6$. What you have written is incorrect. Depending on how much detail your instructors demand, you could also simply say "the statement is obviously true for the base case".

*Starting with $k^2>2k$, you add $2k+1$ to both sides of the inequality to get $k^2+2k+1>2k+2k+1$, which is fine. However, the justification you gave for why you can $1$ to the RHS does not make sense to me. In any case, the following argument is simpler. You want to show that
$$
k^2>2k \implies (k+1)^2>2(k+1) \, ,
$$
so offer the following argument:
$$
k^2+2k+1=(k+1)^2>4k+1>2(k+1) \, .
$$
The final inequality is true because $4k+1>2(k+1) \iff k>-\frac{1}{2}$.

Ex. 2

*

*Your proof is written backwards: rather than showing $k^2>2k \implies (k+1)^2>2(k+1)$, you try to show $(k+1)^2>2(k+1) \implies k^2>2k$. That's not the same thing: note that $x=5 \implies x^2=25$, but $x^2=25{\cancel\implies} x=5$. Sometimes writing proofs backwards is acceptable because all of the steps are reversible. For instance, to prove that for all real numbers $x$ and $y$, $\sqrt{xy}<\frac{x+y}{2}$, we can do the following:
\begin{align}
\sqrt{xy} < \frac{x+y}{2} &\iff 4xy < (x+y)^2 \\[4pt]
&\iff x^2-2xy+y^2 > 0 \\[4pt]
&\iff (x-y)^2>0
\end{align}
Since $\sqrt{xy}<\frac{x+y}{2} \iff (x-y)^2>0$, we can conclude that $(x-y)^2>0 \implies \sqrt{xy} < \frac{x+y}{2}$. Moreover, since the LHS of this implication is true for all real $x$ and $y$. However, when trying to prove that $A \implies B$, this method doesn't work if it is not the case that $B \implies A$. So exercise caution with this method.


*To write a reverse proof by induction, you have to show that $P(k+1) \iff P(k)$. If you start by assuming that $P(k+1)$, you cannot assume that $P(k)$ is true as you did. So your method is not valid.
