Find the smallest possible number of different numbers between $a + b, b + c, c + a, ab + 1, bc + 1, ca + 1, abc$ 
Find the smallest possible number of different real positive number
between $a + b, b + c, c + a, ab + 1, bc + 1, ca + 1, abc$

$a$, $b$, $c$ are real positive numbers and they're all
different. From that,  I can tell that $a+b \ne b+c \ne c+a$, and $ab + 1 \ne bc + 1 \ne ca + 1$, but I couldn't get any further with the exercise. How can I find the answer and prove it?
P. S. I am not very good at these kinds of exercises but couldn't find any good resources. If you know about any, I'd appreciate it if you told me about some.

Consider the inequality chain:
$$ a+b =ab+1 < a+c = ac+1 = abc < b+c < bc+1,$$
which has general solutions $ a = 1,  1 < b < \frac{1+\sqrt{5}}{2} , c = \frac{1}{b-1}$.
As an explicit example, with $a = 1, b = 1.5, c = 2$, we get 4 distinct values: $2.5, 3, 3.5, 4.$
We will show that 4 distinct values is the best we can do.

Proof by contradiction. Suppose we can get 3 distinct values.
WLOG $ a < b < c$.
Notice that $ a+ b < a+c < c + b $, so this must be the 3 distinct values in order.
and $ ab + 1 < ac + 1 < bc + 1$, so this must be the 3 distinct values in order.
Thus, we have $ a + b = ab+ 1, a+c = ac+1, c+b = bc+1$.
This yields $(a-1)(b-1) = 0$, $(a-1)(c-1) = 0$, $(b-1)(c-1) = 0$.
This can be satisfied if and only if (at least) 2 of the variables are equal to 1, which contradicts the assumption that they are distinct.

To come up with the example of 4 distinct values, study the possible chains of intertwining the 2 inequalities, along with $abc$. Several of these lead to solutions. (This is not an exhaustive list)

*

*The initial inequality chain of

$$ a+b =ab+1 < a+c = ac+1 = abc < b+c < bc+1,$$
with solutions $ a = 1,  1 < b < \frac{1+\sqrt{5}}{2} , c = \frac{1}{b-1}$.


*We could have

$$ a+b =ab+1 < a+c = ac+1 < abc = b+c < bc+1,$$
with solutions $ a = 1, 1 < b <2, c = 1 + \frac{1}{ b - 1 } $.


*We could have

$$ a+b = ab+ 1 = abc , a+c, ac+1, b+c = bc+1, $$
which has solutions $0 < a < 1, b=1, c = \frac{a+1}{a}$.
As an explicit example, $ a = \frac{1}{2}, b = 1, c = 3$.


*We could have

$$ a+b < ab+1 = a+c < ac+1 = b+c < bc+1 = abc,$$
with solution $a\approx1.33826, b\approx1.61803, c\approx1.82709$ taken from Wolfram.


*We could have

$$ a+b < ab+1 = a+c =abc < ac+1 = b+c < bc+1,$$
with solution $a\approx 1.27716, b \approx 1.42902, c \approx 1.54792$ taken from Wolfram


*We could have

$$ a+b = abc < ab+1 = a+c < ac+1 = b+c < bc+1,$$
with solution $a\approx 1.26523, b\approx 1.39919, c\approx 1.50507$ taken from Wolfram.
