Prove that there do not exist three distinct real number a,b and c such that all of the numbers a + b + c, ab, ac, bc, abc are equal.

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Can you guys tell me if this is right?

Assume, to the contrary, there exist three distinct real number a,b and c such that all of the numbers a + b + c, ab, ac, bc, abc are equal. Then $4(a+b+c)=abc + ab + ac + bc$. Since, $a(4-b) + b(4-c) + c(4-a) - abc - abc + abc=0$, it follow that $a(4-b-bc) + b(4-c-ac) + c(4-a + ab) = 0$. Since $a(0) + b(0) + c(0) = 0$, it follows that $4=b+bc$ and $4=c+ac$ and $4=a-ab$. As a result, $b=\frac{4}{1+a}$ and $c=\frac{4}{1+c}$ and $a=\frac{4}{1-b}$. Hence, $a$,$b$ and $c$ are 3 distinct real number which is a contradiction.

You do not even to include $a+b+c$ or $abc$, as there are no distinct $a,b,c$ such that $ab=ac=bc$. — Varun Vejalla, Jul 10 '19 at 17:54
Why does $a(4-b-bc) + b(4-c-ac) + c(4-a + ab) = 0$ imply that $4=b+bc$ etc? — Martin R, Jul 10 '19 at 17:57
Btw, do you mean “pairwise distinct numbers” or “not all equal numbers”? — Martin R, Jul 10 '19 at 17:58

score 2 · Answer 1 · answered Jul 10 '19 at 21:15

Suppose that $ab=bc=ac$ for some real $a,b,c$.

Case 1: At least one of $a,b,c$ is zero.

Wlog. $a = 0$. Then $bc=0$, so either $b=0$ or $c=0$. So two of our numbers are equal.

Case 2: None are zero. So divide $ab=ac$ by $a$, giving that $b=c$.

In either case, at least two are equal, as required.

Prove that there do not exist three distinct real number a,b and c such that all of the numbers a + b + c, ab, ac, bc, abc are equal.

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