Given $n > 1$, we consider the finite sets of positive integers which sum to $n$, and out of these sets we want to maximize the product. For example, given $n = 6$, the set $\{1, 5\}$ does not maximize the product, while the set $\{3, 3\}$ does.
How do I prove that for any given $n$, the set which maximizes the product does not contain any element greater than or equal to $4$?
I've started off with the basic idea that given $n$ is the sum, we can break $n$ up into the parts: $(n - x) + x$, where $x$ is some number between $0$ and $n$, non-inclusive of both.
Then the product becomes $(n - x)x$, and to maximize this function we take the derivative, find the critical point and find that this function is maximized when $x = \frac{n}{2}$.
But I'm not sure how to continue this process to prove that the greatest element in the set that maximizes the product is less than or equal to $4$?
Because my current attempt, for the example of $n = 12$, seems to yield that the maximum product is $6 \times 6$, not $3 \times 3 \times 3 \times 3$.
Suppose that our partition contains an entry $m\gt 4$. We show that it is profitable to split $m$.
It is convenient to divide into two cases, $m$ odd and $m$ even.
Suppose that $m$ is odd, say $m=2k+1$. We show that it is profitable to split $m$ as $k+(k+1)$. It is enough to show that $k(k+1)\gt 2k+1$, or equivalently that $k^2-k-1\gt 0$. Since $k\ge 2$, we have $k^2-k-1\ge 2k-k-1=k-1\gt 0$.
Suppose that $m$ is even, say $m=2k$. We show that it is profitable to split $m$ as $k+k$. It is enough to show that $k^2\gt 2k$, that is, that $k(k-2)\gt 0$. Since $k\ge 3$, this inequality holds.
As has been pointed out by Daniel Fischer, the case $m=4$ is special. We can either keep or split as $2+2$ without changing the product.