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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.

vlnkaowo
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    I don't know that there are a lot of "these kind of exercises"—this one looks pretty ad hoc. Hint: taking one of the variables equal to $1$ should result in the possibility of having only four distinct values, which I believe is the least possible (using the non-equalities you mentioned), although I don't know a quick proof. – Greg Martin Nov 25 '20 at 08:35
  • @GregMartin I am not sure either, that's why I put it just as a little footnote. I've got a bunch of similar exercises (for example the same rules for $a+2b, b+2c, c+2a$) and thought there could be a parent name for them. Going with $1$ seems to be the way, I just need to put it in a "legit" way – vlnkaowo Nov 25 '20 at 08:41

1 Answers1

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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)

  1. 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}$.

  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 } $.

  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$.

  1. 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.

  1. 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

  1. 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.

Calvin Lin
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