Finding minimal and maximal value of $a+b+c$

Question

$a$, $b$ and $c$ are $\mathbb Z^+$ with different numbers.

$abc = 40$

• What is the maximal value of $a+b+c$?

• What is the minimal value of $a+b+c$?

I'm wondering what the most useful way to solve is. By the way, I wanted to ask you. I mean If there's a way that is always usable, let me know.

Regards

Answer 1

By AM-GM \begin{eqnarray*} \frac{a+b+c}{3} \geq \sqrt[3]{abc} \end{eqnarray*} So $a+b+c \geq 3 \sqrt[3]{40}= 10.3 \cdots$, so the minimal value that could attained is $11$ and this is done with $a=5,b=4,c=2$.

The maximum is attained when $a=1,b=2,c=20$ and is $23$.

Answer 2

for b) using the AM-GM inequality $$a+b+c\geq 3\sqrt[3]{abc}$$ so we have $$a+b+c\geq 3\sqrt[3]{40}$$

Answer 3

To find the minimum you can use the AM-GM inequality:

$$a+b+c \ge 3\sqrt[3]{abc}\,,$$

with equality holding iff $a=b=c$. In this instance you have

$$3\sqrt[3]{abc} \approx 10,25$$

so that the minimum value you can obtain is greater or equal than $11$. Easy to see that $a=2, b=4, c=5$ gets you to $11$.

Answer 4

40 is not big enough to avoid going through all factorizations: $$ \begin{split} 40 &= 40 * 1 * 1 = 20 * 2*1 = 10 * 4 * 1 \\ &= 10 * 2 * 2 = 5 * 8 *1 = 5*4*2 \end{split} $$ so the sums are $42^*, 23, 15, 14^*, 14, 11$, where numbers coming from factorization with duplicates are marked by ${}^*$.