To a certain extent, a proof by induction does LOOK like a bit of fancy rewriting.
Let's proceed with a bit more formality and try a different route.
32−7(3)+12=9−21+12=−12+12=0. You did this step (but with a typographical error) in your first post.
Then you loosely expressed the second step (although n cannot be both k and k + 1).
k is an integer≥3 such that k2−7k+12=u≥0⟹
(k+1)2−7(k+1)+12=k2+2k+1−7k−7+12=(k2−7k+12)+2k+1−7=u+2k−6≥0+6−6=0.
THEREFORE n2−7n+12≥0 for every integer ≥3.
We just rewrote
(k+1)2−7(k+1)+12 in terms of
k2−7k+12 plus some adjustments.
Then we show that the adjustments do not adversely affect what we assumed to be true of k.
The INTUITION is this.
If we show that something is true of some particular integer and also show that, if it is true for any specific integer, then it is also true for the next higher integer, the whole remaining row of dominoes falls. It is true of 3, so it is true for 4, but that means it is true for 5, which entails that it is true for 6, and so on world without end.
The easy part is almost always showing that it is true for some integer.
The tricky part is showing that if it is true for some arbitrary integer it is also true for the next integer. That part of the proof almost always comes down to expressing the property of interest for the next integer in terms of the arbitrary integer.
Does this help or make things worse?