This is a question I originally asked on MSE, receiving no answer, even with a bounty (which expired) on it. Therefore I am crosslinking in order to prevent duplication of effort: see [here](http://math.stackexchange.com/questions/1056240/two-strictly-related-proofs-by-induction-of-inequalities) for the original question.


Predictably, I am stuck with the inductive steps.
Let $a_n=\prod_{i=1}^m p_i^{b_i}$, where $b_1=10;b_2=6;b_3,b_4=4;b_5,b_6=3;b_7,\ldots ,b_{n+1}=2$ and set $\lim_{n,m\to \infty}\frac{\log a_n}{p_m}=1$. Suppose also this ratio converges to $1$ faster than $\displaystyle\frac{p_{m+1}}{p_m}$, so that if $n$ is large enough, we always have $\log a_n<p_{m+1}$. 

I want to prove that for sufficiently large $n$, with $c$ being a constant and $q<m$, if $$\frac{c}{\log \log a_n}<\frac{\left(1+{\prod_{i=1}^m\left(p_i^{b_i+1}-1\right)^{1/m}}\right)^m}{\prod_{i=1}^m\left(p_i^{b_i+1}-1\right)}, \tag{1}$$ then the following statements are true: 
>$$ \frac{c}{\log \left(\log a_n+\log p_q\right)}<\\\frac{\left(1+\prod_{i=1}^{q-1}\left(p_i^{b_i+1}-1\right)^{1/m}\cdot\left(p_q^{b_q+2}-1\right)^{1/m}\cdot\prod_{i=q+1}^{m}\left(p_i^{b_i+1}-1\right)^{1/m}\right)^m}{\prod_{i=1}^{q-1}\left(p_i^{b_i+1}-1\right)\cdot\left(p_q^{b_q+2}-1\right)\cdot\prod_{i=q+1}^{m}\left(p_i^{b_i+1}-1\right)}; \tag{2}$$



>$$ \frac{p_{m+1}}{p_{m+1}-1}\frac{c}{\log \left(\log a_n+\log p_{m+1}\right)}<\frac{\left(1+\prod_{i=1}^{m+1}\left(p_i^{b_i+1}-1\right)^{1/(m+1)}\right)^{m+1}}{\prod_{i=1}^{m+1}\left(p_i^{b_i+1}-1\right)}. \tag{3}$$

To clear it up, in $(2)$ we have $a_n\cdot p_q=a_{n+1}$, in $(3)$ instead $a_n\cdot p_{m+1}=a_{n+1}$.

$(2)$ is fairly intuitive, as the LHS goes to $0$ as $n\to \infty$ while the RHS goes to $1$, but that doesn't tell me so much since if the former is larger than $1$ and slightly smaller than the latter, I cannot say *a priori* that the LHS is sufficiently fast in its convergence to $0$, to be always less than the RHS.
On the other hand, it is only my istinct that says $(3)$ holds, but I might be wrong.

Here is how I tackled both inequalities, hoping to simplify things a bit (and not "too much"). Call $L_t$ and $R_t$ respectively the LHS and RHS of $(1)$, $(2)$ and $(3)$. So $(2)$ is the same as $$ L_1 \frac{L_2}{L_1}<R_1\frac{R_2}{R_1},$$ and since $L_1<R_1$ by hypothesis, $(2)$ is implied by $$ \frac{\log \log a_n}{\log \left(\log a_n+\log p_q\right)}<\\ \frac{p_q^{b_q+1}-1}{p_q^{b_q+2}-1}\left(\frac{1+\prod_{i=1}^{q-1}\left(p_i^{b_i+1}-1\right)^{1/m}\cdot\left(p_q^{b_q+2}-1\right)^{1/m}\cdot\prod_{i=q+1}^{m}\left(p_i^{b_i+1}-1\right)^{1/m}}{1+{\prod_{i=1}^m\left(p_i^{b_i+1}-1\right)^{1/m}}}\right)^m.\tag{4}$$
Similarly, $(3)$ follows from $$ \frac{p_{m+1}}{p_{m+1}-1}\frac{\log \log a_n}{\log \left(\log a_n+\log p_{m+1}\right)}<\\ \left(1+\prod_{i=1}^{m+1}\left(p_i^{b_i+1}-1\right)^{1/(m+1)}\right)\left(\frac{1+\prod_{i=1}^{m+1}\left(p_i^{b_i+1}-1\right)^{1/(m+1)}}{1+{\prod_{i=1}^m\left(p_i^{b_i+1}-1\right)^{1/m}}}\right)^m.\tag{5}$$
This said, I do not know how to prove $(4)$ and $(5)$ either. Any ideas? Thanks in advance.

**EDIT**: I have succeeded in making sufficient to prove the inequalities having $p_{m+1}-1$ instead of $p_m$, which somewhat might be a tiny bit easier.