Confusion about this math proof

I had to prove that if a and b are two positive integers then a^2(b+1)+b^2(a+1)≥4ab.

The proof that I saw goes like this:
(a-b)^2=a^2-2ab+b^2≥0

a^2+b^2≥2ab .....1

2a^2+2b^2≥4ab

Becase a and b are two positive integers a≥1 and b≥0

Therefore
a^2(b+1)+b^2(a+1)≥4ab= a^2(1+1)+b^2(1+1)≥4ab. (from 1)
=2a^2+2b^2≥4ab
Thus
a^2(b+1)+b^2(a+1)≥4ab

The reason why I am confused is that in this solution they assign a and b the number 1.This is something I haven't seen before. Is this a correct way?

So we know
(a-b)^2 >= 0
from this we can conclude
2a^2 + 2b^2 >= 4ab

Now we are attempting to prove that
a^2(b+1) + b^2(a+1) >= 4ab
and we know that b>=1 and a>=1
so the minimum values that b and a can have are 1.
and we also know that
a^2 * 2 + b^2 * 2 >= 4ab
is a true statement because of what we showed at the very beginning
so clearly if we give a and b their minimum values and the statement is true then the statement will definitely be true for any values greater than those.
Hence we give a and b their minimum values and get
2a^2 + 2b^2 >= 4ab which is true
if a or b were greater than 1 then we would get
Xa^2 + Yb^2 >= 4ab
and we know X and Y would both be greater than 2
so of course that statement would be true.

integers are whole numbers
if a=0 and b=0 then there is 0≥0
nothing to work with
so the next best positive thing is 1
if a=0 and b≥1 then 2ab still =0
yu still have not much to work with
so it is ok to asume both a and b ≥1 other then 0