supA and infA, if $A:=\{\frac {a+b+c}{abc}: a,b,c \in \Bbb N\}$

Question

Find supA, and infA, if $A:=\{\frac {a+b+c}{abc}: a,b,c \in \Bbb N\}$

I think $supA=3$ and $infA=0$

Am I right?

Answer 1

$a,b,c > 0$ so $a + b + c > 0$ and $abc > 0$ and $\frac {a+b+c}{abc} > 0$ so $0$ is a lower bound.

But so is $-27\frac 35$..... To prove that $0 = \inf A$ is the greatest lower bound we must show that for any $k > 0$ we can find a $\frac {a+b+c}{abc} < k$ so that $k > 0$ can not be a lower bound.

That's not hard if $a = b=c$ (just to make things easy) we need $\frac {3a}{a^3} < k$ or $\frac 3k < a^2$ or simple choose any $a > \sqrt{\frac 3k}$.

$\frac {a+b+c}{abc} =\frac a{abc} + \frac b{abc} + \frac c{abc} = \frac 1{bc} + \frac 1{ac} + \frac 1{bc} \le 1+1+1 = 3$. So $3$ is an upper bound.

If $a = b = c =1$ we get $\frac {a+b+c}{abc} =3\in A$ so nothing less than $3$ can be an upper bound. So $3$ is the least upper bound and $3 =\sup A$.

Answer 2

We know that $0 \leq \inf A$. Let $c := b := a$ for arbitrary $a \in \mathbb{N}$, then $\dfrac{a+b+c}{abc} = \frac{3a}{a^3} = \frac{3}{a^2}$. As $a \rightarrow\infty$, this fraction goes to zero, hence $\inf A\leq 0$.

To determine $\sup A$, try to prove that $\dfrac{a+b+c}{abc} \leq 3$ for all $a,b,c \in \mathbb{N}$.