Bound for the sum of the divisors of a number 
Let us denote by $s(n) = \sum_{d|n} d$ the sum of divisors of a natural number $n$ ($1$ and $n$ included). If $n$ has at most $5$ distinct prime divisors, prove that $s(n) < \dfrac{77}{16}n$. Also prove that there exists a natural number $n$ for which $s(n) > \dfrac{76}{16}n$ holds.

Attempt:
Let $n = p_1^{\alpha_1}p_2^{\alpha_2}p_3^{\alpha_3}p_4^{\alpha_4}p_5^{\alpha_5}$ where the $p_i$ are primes. The sum of the divisors is $$(1+p_1+\cdots+p_1^{\alpha_1})(1+p_2+\cdots+p_2^{\alpha_2})\cdots(1+p_5+\cdots+p_5^{\alpha_5})=\frac{(p_1^{\alpha_1+1}-1)(p_2^{\alpha_2+1}-1)\cdots(p_5^{\alpha_5+1}-1)}{(p_1-1)(p_2-1)\cdots(p_5-1)}.$$ Then we see that $\dfrac{s(n)}{n}=\prod_{i=1}^5 \dfrac{p_i^{\alpha_i+1}-1}{p_i^{\alpha_i}(p_i-1)}$. But then $$\dfrac{p_i^{\alpha_i+1}-1}{p^\alpha_i(p_i-1)} = \dfrac{p_i-\dfrac{1}{p_i^{\alpha_i}}}{p_i-1}<\dfrac{p_i}{p_i-1}=1+\dfrac{1}{p_i-1} = \dfrac{p_i}{p_i-1}.$$ Since the last expression is decreasing for all $p_i$, we have
$$\prod_{i=1}^5 \frac{p_i^{\alpha_i+1}}{p_i^{\alpha_i}(p_i-1)} < \prod_{i=1}^5 \dfrac{p_i}{p_i-1} \le \dfrac{2}{1}\cdot \dfrac{3}{2}\cdot \dfrac{5}{3}\cdot \dfrac{7}{6}\cdot \dfrac{11}{10}=\dfrac{77}{16}.$$
If we have less than $5$ primes, notice that by repeating the same process we arrive at smaller bounds. If $n = 1$ then clearly our bound holds.
How do we find an $n$ such that $s(n) > \dfrac{76}{16}n$?
 A: You have proven that $n$ has to have at least $6$ distinct prime divisors. From your work it should be clear that for prime $p$, $\frac {s(p^n)}{p^n}$ is decreasing with $n$, so you want $n$ to be the product of the first powers of primes.  The natural choice is to keep multiplying primes $p_i$ until $\prod \frac {p_i+1}{p_i} \gt \frac {76}{16}$ Alpha finds the product up to $i=20$ to be generous:  $ \prod_{i=1}^{20} \frac {p_i+1}{p_i} \gt 4.769\gt \frac {76}{16}$
A: Let $n = \prod_i^6 p_i^k$.
You've pointed out that $s(n)/n = \prod_i^6 \frac{p_i - \frac 1{p_i^k}}{p_i -1} < \prod_i^6 \frac{p_i }{p_i -1}=77*13/16*12$.
But we can make $k$ arbitrarily large so for any $\epsilon > 0$ in particular any epsilon $77/16 < 77*13/16*12 - \epsilon$ we can find $k$ so that $77*13/16*12 >s(n)/n = \prod_i^6 \frac{p_i - \frac 1{p_i^k}}{p_i -1} > 77*13/16*12 - \epsilon$.
I figure, but haven't varified, that as 
$\prod_i^6\frac{p_i}{p_i - 1} = 77*13/16*12 = 1001/192=5.213$
$\prod_i^6\frac{1 + p_i}{p_i} = 32256/30030= 1.0741$
$\log_{1.0741} 5.213 = 23.01$
that $k = 24$ ought to be large enough the $n = \prod_i^6 p_i^{24}$ will satisfy.
Not sure if my reasoning for $k =24$ is valid but if not some larger $k$ will do.
