For $n \in \mathbb{Z} : n \geq 1$
$ f(n) = \displaystyle\max_{\substack{ x_1+\dotsm+x_k = n\\ x_i\in\mathbb{Z}^{+} }} x_1 x_2 \dotsm x_k $
$ f(n) = \begin{cases} 1 & \text{if $n = 1$}, \\ 3^{\left\lfloor\frac{n}{3}\right\rfloor}& \text{if $n \mod 3 = 0$},\\ 3^{\left\lfloor\frac{n}{3}\right\rfloor-1} \cdot 2^2& \text{if $n \mod 3 = 1$},\\ 3^{\left\lfloor\frac{n}{3}\right\rfloor} \cdot 2& \text{if $n \mod 3 = 2$}, \end{cases} $
Proof:
First observe that for any $x_i = 1$ in our product we do not increase the value of the final product. Therefore we want $x_i > 1$. However, for any set of three $x_i = 2$ we want to refactor this set into two $x_i = 3$ since $3\cdot3 > 2\cdot2\cdot2$. Now, for any $x_i > 4$ observe that $2(x_i-2) = 2x_i - 4 > x_i$. Thus to maximize our product we have at most two 2's and as many 3's as possible.
What is my proof possibly missing?
Here's an updated proof.
Proof:
First observe that for any $x_i = 1$ in our product we do not increase the value of the final product. Therefore we want $x_i > 1$. However, for any set of three $x_i = 2$ we want to refactor this set into two $x_i = 3$ since $3*3 > 2*2*2$. Now, for any $x_i > 4$ observe that subtracting 2 from our term increases our total product. That is $2(x_i-2) = 2x_i - 4 > x_i$ for $x_i > 4$. We can repeat this process until $x_i \leq 4$ and this limits our individual factors to being no greater than 4. For any $x_i = 4$ where there are no existing factors equal to 2 we do not necessarily need to factor this into $2 \cdot 2$ since this product is equal to $4$ and will not increase our total product. Thus to maximize our product we have at most two 2's, or a single 4, and as many 3's as possible. This leads to our piecewise defined function. For any n divisible by 3 we will have a product comprised of 3 raised to the power $\frac{n}{3}$. If we have a remainder of 1 when dividing n by 3 we want one less factor of 3 than $\frac{n}{3}$ provides so that we can create a factor of $2^2$ or $4$. Finally, if n divided by 3 has a remainder of 2 we will just include the remainder as a factor in our final product.