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

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.

• Isn't it easier to rewrite the inequality as $$\frac1a+\frac1b+\frac1c>1?$$ Then one of $1/a$, $1/b$, $1/c$ must be $>1/3$. – Angina Seng May 8 '20 at 3:27
• That will be $ab+bc+ca≤3bc≤abc$ as $a≥3$. – Alapan Das May 8 '20 at 3:32
• I don't understand why, for example, $ab + bc + ca \le 2bc$ isn't true, though, if we assume $a \ge 2$. – David Dong May 8 '20 at 3:35
• $ab\le bc,ac\le bc,bc\le bc$ because $a\le b\le c$. Then add them up. – Empy2 May 8 '20 at 3:42
• @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! – healynr May 8 '20 at 3:42

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.

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.

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

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 $$x,y,z \in \mathbb N$$.