Induction on more than one variable I've been reading about induction on many variables and would like to check if I am understanding it the right way.
Problem: Let $n, m \ge 1$ integers. Prove: $nm$ $\ge$ $n + m - 3$ for every $n, m \in$ $\mathbb N$
Option 1: Induct on $n$ leaving $m$ fixed (arbitrary) which, for me, is not inducting on two variables but proves the statement:

*

*Base case: let $k = 1$, then $m \ge m - 2$

*Induction hypothesis: $nm$ $\ge$ $n + m - 3$

*Let $ k = n + 1$. Then  $(n+1)m = nm + m $ $\ge$ $n + m - 3 + m$ $\ge$ $n + m - 3 + 1 = (n+1) + m - 3$
where I used IH on the first $"\ge"$ and the fact that m $\ge$ 1 on the following one.

As $m$ is arbitrary in $\mathbb N$ induction proves the statement for all $n, m$.
Option 2: Two variable induction:

*

*Base case: let $k = 1, q = 1$, then $1\cdot1 = 1 \ge 1 + 1 - 3 = -1$

*Induction hypothesis: $nm$ $\ge$ $n + m - 3$

*Let $ k = n + 1, q= m$. Then  $(n+1)m = nm + m $ $\ge$ $n+m-3 + m \ge n+m-3 + 1 = (n+1)+m-3 $

*Let $ k = n, q = m + 1$. Then $n(m+1) = nm + n $ $\ge$ $n+m-3 + n \ge n+m-3 + 1 = n+(m+1)-3 $
Proving base case, $(n+1, m)$ and $(n, m+1)$ ends the proof.
Are both okay? Thanks :)
 A: 
Is $nm \geq n + m - 3 ~: ~n,m \in \Bbb{Z^+}$?

3rd option.  Nested induction. 
For illustrative purposes only. 
Conjecture could clearly be more easily proven directly.
Is the conjecture true when $m = 1$?
Is $n \geq n + 1 - 3$?  
For $n = 1$ : true : because $1 \geq 1 + 1 - 3.$
Assume true for $n = N$.
Then $(N+1)(1) = N(1) + 1 \geq (N + 1 - 3) + 1.$ 
[Because of the induction assumption].
But $(N + 1 - 3) + 1$ can be rewritten 
$[N+1] + 1 - 3.$ 
Thus, $(N+1)(1) \geq [N+1] + 1 - 3.$
Consequently, when $M = 1$, the conjecture has been proven to hold for all positive integers $(n)$.

Assume the conjecture is true for $m = M$. 
Here, the inductive assumption is being made that 
when $m = M,$ the conjecture holds for any positive integer $n$.
Is the conjecture true for $m = M+1$?
With $m = M+1$, suppose $n = 1.$
Is $1[M+1] \geq 1 + [M+1] - 3$?
The LHS is $M+1$ and the RHS is $M+1 - 2$, so the answer is Yes.
So, now, with $m = M+1$, the conjecture has been proven for $n = 1.$
With $m = M+1$, assume that the conjecture has been proven for $n = N$.
Is the conjecture true for $n = [N+1]$?
$[N+1][M+1] = N[M+1] + [M+1] \geq $ 
[by inductive assumption] 
$(N + [M+1] - 3) + [M+1]) = [N+1] + [M+1] - 3 + M$. 
This is $ ~~> [N+1] + [M+1] - 3.$ 
Therefore, $[N+1][M+1] \geq [N+1] + [M+1] - 3.$
Therefore, when $m = M+1$, the conjecture has been proven true for both $n=1$ and $n = N+1$.
Therefore, the conjecture is true for $m = M+1$.

Thus, the conjecture is true for any positive integer $n$, when $m = 1.$
Also, the conjecture is true for any positive integer $n$,
when $m = M+1$.
Therefore, the conjecture is true for any positive integer $n$ and any positive integer $m.$
