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Problem statement (from Andreescu and Andrica NT: SEP):

Find all positive integers $a, b, c$ such that $ab + bc + ac > abc$.

The solution starts with this:

Assume that $a \le b \le c$. If $a \ge 3$, then $ab + bc + ac \le 3abc \le abc$, a contradiction.

I understand why this would be a contradiction, of course, but I don't understand how the inequality works, nor what is special about $a$ being greater than $3$ rather than another number. In particular, I don't understand why $ab + bc + ac \le 3bc$. For example, why is it not true that $ab + bc +ac \le 2bc$ if we assume $a \ge 2$? I see that if $a \ge 3$, then clearly $3 \le b \le c$, which might be important for understanding the solution because now the inequality flips, but I'm still confused. I've tried to understand this solution better, but I still don't understand it.

I also see that a similar problem has been answered here: Find all prime $a, b, c$ such that $ab+bc+ac > abc$, but while it uses a similar method, it doesn't answer my question.

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    $\begingroup$ Isn't it easier to rewrite the inequality as $$\frac1a+\frac1b+\frac1c>1?$$ Then one of $1/a$, $1/b$, $1/c$ must be $>1/3$. $\endgroup$ – Angina Seng May 8 '20 at 3:27
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    $\begingroup$ That will be $ab+bc+ca≤3bc≤abc$ as $a≥3$. $\endgroup$ – Alapan Das May 8 '20 at 3:32
  • $\begingroup$ I don't understand why, for example, $ab + bc + ca \le 2bc$ isn't true, though, if we assume $a \ge 2$. $\endgroup$ – David Dong May 8 '20 at 3:35
  • $\begingroup$ $ab\le bc,ac\le bc,bc\le bc$ because $a\le b\le c$. Then add them up. $\endgroup$ – Empy2 May 8 '20 at 3:42
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    $\begingroup$ @DavidDong $ab+bc+ac > abc \implies \frac{1}{c}+\frac{1}{a}+\frac{1}{b}>1$. So say $a \le 2$, then $\frac{1}{a} \le \frac{1}{2}$. So $\frac{1}{c}+\frac{1}{b} \ge \frac{1}{2}$, which is entirely possible!. But if you do $a \le 3$, then you would need $\frac{1}{c}+\frac{1}{b} \ge \frac{2}{2}$, which is not possible! $\endgroup$ – healynr May 8 '20 at 3:42
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It is a typo. It should say that if $a \ge 3$, we have $ab+ac+bc \le 3bc$ because $ab, ac \le bc$. Then $3bc \le abc$ because we assumed $a \ge 3$. Now we have $ab+ac+bc \ge abc$ which contradicts what we want to find.

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The contradiction demonstrates that $a$ cannot be greater than or equal to $3$. Thus we know the least element of $a,b,c$ must be $2$ or less. Since $a,b,c$ are positive integers, we know $a = 1,2$. If you plug in $a=1$, you'll quickly notice a contradiction, so you can conclude $a=2$ and go from there.

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$ab+bc+ca>abc \rightarrow c(a+b)>ab(c-1) \rightarrow (a+b)(1-\frac{1}{c})=ab-\frac{ab}{c}$

Now, if we can find all $a,b$ such that $a+b>ab$ then we will bound the problem.

W.L.O.G $a≥b \Rightarrow ab≥2a≥a+b$ for $b≤2$. So, we have bounded.

So, 1. For $b=1$ $a+1≥ab=a$

  1. For $b=2$ $a+2≥2a \rightarrow a=2$(as we took $a≥b$)

So, for $b=1$ for all $a,c>0$ $ab+bc+ca≥ac$ and for $b=2,a=2$ this becomes $4+4c>4c$, so satisfied for all $c$.

So, the triplet will be permutations of $(x,1,y);(2,2,z)$ for $x,y,z \in \mathbb N$.

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