Let $d(n)$ be the number of divisors function, i.e., $d(n)=\sum_{k\mid n} 1$ of the positive integer $n$.
I am interested in estimating, the following sum $$ A(a,x)=\sum_{n\leq x} \min[ d(n), M]^a $$ from below for some function $M$ depending on $x,$ the upper limit.
For now, let $a=1.$
In the answer to the following question where the case $a=1$ was considered, the lower bound on $A(a,x)$ for the case $M\leq (\log x)^{10},$ was stated to be of the same order as that of the upper bound in that answer.
Reading over the answer much later now, it is not so clear to me what the lower bound actually is, because the upper bound is also not fully stated. I have looked at the Selberg-Delange chapter, and the subsequent chapter in Tenenbaum's Introduction to Probabilistic and Analytic Number Theory book as suggested, but it is still not fully clear what's going on.
For example, Theorem 4, p. 205 in that book states $$ \pi_k(x)=\frac{x}{\log x}\frac{(\log \log x)^{k-1}}{k!}\left\{\lambda\left( \frac{k-1}{\log\log x}\right)+ O\left(\frac{k}{(\log \log x)^2}\right)\right\} $$ where $$ \pi_k(x)=\mid \{ n\leq x: \omega(n)=k \} \mid,$$ and $$\lambda(z)=\frac{1}{\Gamma(z+1)}\prod_p \left(1+\frac{z}{p-1}\right)\left(1-\frac{1}{p}\right)^z.$$ and $k$ is allowed to grow with $n,$ with $1\leq k\leq A \log \log x.$
So presumably I need an estimate of the form $$ \sum_{1\leq k:2^k\leq M} 2^k \pi_k(x), $$ for the lower bound. What are the explicit steps in establishing this?
