Prove that product of 3 numbers with a fixed sum is highest when they are equal

Question

How to prove the fact, that product of any three numbers with a fixed sum is highest possible when they equal $sum/3$, meaning they are equal?

Example:

Let the "sum" be a constant $K$ and the "numbers" $a, b, c$.

given $K = 3$
How can I prove that $a,b,c$ make the highest product when they equal $K/3 = 1$?

Thanks in advance!

Answer 1

Set $$a+b+c=s$$ then we have $$\frac{a+b+c}{3}\geq \sqrt[3]{abc}$$ if $$a,b,c$$ are assumed to be positive. This is the $$AM-GM$$ inequality. The equal sign holds if $$a=b=c$$

Answer 2

It's for the same reason area and perimeter ratio is highest at a circle; its best to use the least perimeter (lowest sum of numbers),as this means we can get the most area for our perimeter. Thus it is best to use the same number thrice.