number theory - Prove that product of five divisors is $leq n^4$

Five divisors sum $(n_1+n_2+n_3+n_4+n_5)$ of natural number $n>0$ is prime number. Prove that the product of these five divisors is $\leq n^4$
in math words $n_1\cdot n_2\cdot n_3\cdot n_4\cdot n_5 \leq n^4$.
And that's what I did. Let's say that

$n_1.

Let $n_5$ is biggest divisor of number n $\implies n_5=n$

We want to find other four biggest divisors. It obvious can be $\frac{n}{2},\frac{n}{3},\frac{n}{4},\frac{n}{5}$.

Multiplication of these divisors would be $n\cdot \frac{n}{2}\cdot \frac{n}{3}\cdot \frac{n}{4}\cdot\frac{n}{5}=\frac{n^5}{120}$

Now we have inequality $\frac{n^5}{120}\leq n^4$.

Since $n>0$ so lets divide by $n^4$

$\implies \frac{n}{120}\leq1$ or $n\leq120$

So I got that number $n$ must be smaller than $120$. But how to prove it for bigger numbers? Thanks in advance.

Answer

It's not specific to $5$, it works for every $m \geqslant 2$. If $n$ is a positive integer, and $n_1, \dotsc, n_m$ are — not necessarily distinct — positive divisors of $n$ such that
$$\sum_{k = 1}^m n_k \quad\text{ is prime,} \tag{1}$$
then
$$\prod_{k = 1}^m n_k \leqslant n^{m-1}. \tag{2}$$

Writing $n_k = n/d_k$, i.e. setting $d_k = n/n_k$, for $1 \leqslant k \leqslant m$, the inequality $(2)$ is equivalent to
$$\prod_{k = 1}^m d_k \geqslant n\,. \tag{3}$$

Now, the assumption $(1)$ implies $\gcd(n_1, \dotsc, n_m) = 1$. And
$$\gcd(n_1,\dotsc, n_m) = \gcd\biggl(\frac{n}{d_1},\dotsc, \frac{n}{d_m}\biggr) = \frac{n}{\operatorname{lcm}(d_1,\dotsc, d_m)}\,,$$
so we have
$$n = \operatorname{lcm}(d_1, \dotsc, d_m) \leqslant \prod_{k = 1}^m d_k\,,$$
which is $(3)$.