Is the sum always bigger than $n^2$? Let $s(n)$ an arithmetical function defined as
$$s(n)=(p_1+1)^{e_1} (p_2+1)^{e_2} \cdots (p_m+1)^{e_m}$$
where prime factorization of $n$ is $n=p_1^ {e_1} p_2 ^{e_2} \cdots p_m^{e_m}$.
(For example, $s(49)=s(7^2)=8^2=64$, $s(60)=s(2^2 \times 3^1 \times 5^1)=3^2 \times 4^1 \times 6^1=216$)

Prove or disprove the following proposition;
$\phantom{a}\bullet$ There exists a positive integer $M$ such that
$$\forall N>M(N\in\mathbb{N})\phantom{;}; \phantom{;}\sum_{n= 1}^{N}s(n)>N^2$$
 A: This is true.  If $\nu_p(n)$ is maximal such that $p^{\nu_p(n)}\mid n$, then
$$
s(n)\ge (\frac32)^{\nu_2(n)} (\frac43)^{\nu_3(n)}n.
$$
Therefore, for all $i=1$, $\dots$, $2^5 3^3$,
$$
s(2^5 3^3n+i)\ge (\frac32)^{\min(\nu_2(i),5)} (\frac43)^{\min(\nu_3(i),3)} (2^5 3^3 n+i).
$$
Summing over $i=1$, $\dots$, $2^5 3^3$ gives
\begin{eqnarray*}
\sum_{1\le i\le 2^5 3^3} s(2^5 3^3n+i)&\ge&1555910n+ 785732\\
&\ge& 2.08 \sum_{1\le i\le 2^5 3^3}( 2^5 3^3n + i).
\end{eqnarray*}
This proves that
$$
\sum_{1\le n\le N} s(n)\ge 2.08 \frac{N(N+1)}{2}>N^2
$$
whenever $N$ is a multiple of $2^5 3^3=864$.  If $N$ is not a multiple of $864$, write $N=N'+N''$, where $N'$ is a multiple of $864$ and $1\le N''\le 863$.  Then, since $s(n)\ge n$ for all $n$,
$$
\sum_{1\le n\le N} s(n)\ge 2.08 \frac{N'(N'+1)}{2} + N' N'' + \frac{N''(N''+1)}{2}
$$
which will be bigger than
$$
N^2 = N'^2 + 2 N' N'' + N''^2
$$
whenever $N'\ge 22464$.  Also, computation shows that
$$
\sum_{1\le n\le N} s(n)> N^2
$$
whenever $24\le N\le 22463$.  So, we can take $M=23$.
Asymptotically, $s(n)/n$ should behave similarly to the random variable
$$
W:=\prod_p (1+\frac{1}{p})^{Z_p-1},
$$
where the $Z_p$'s are independent geometrically distributed random variables with success probability $1-\frac{1}{p}$, and 
$$
{\Bbb E}(W) = \prod_p \frac{p(p-1)}{p^2-p-1} \approx 2.67411.
$$
