A proof of a statement deduced from Firoozbakht's conjecture and the definition of the sum of divisors function, and an optional question

Question

It is easy to prove that Firoozbakht's conjecture implies (for $n\geq 2$ is equivalent to) $$\sigma(p_n^n)> \frac{p_{n+1}-1}{p_n-1}\sigma(p_{n+1}^{n-1})\tag{1}$$ $\forall n\geq 2$, where $\sigma(m)=\sum_{d\mid m}d$ denotes the sum of divisors function. See if you need the Wikipedia's article dedicated to Firoozbakht's conjecture.

With easy calculations I've deduced from $(1)$ an easy statement, and I would like to know if it is right.

Question 1. Provide us a proof, or the proof for a similar statement: For each integer $N\geq 2$ one has $$\sigma\left(\prod_{n=1}^N p_n^n\right)> \frac{3}{2}(p_{N+1}-1)\prod_{n=2}^N\sigma(p_{n+1}^{n-1}),\tag{2} $$ on assumption of Firoozbakht's conjecture. Many thanks.

This is an optional question that you can answer if you want, with the purpose to obtain feedback about my previous approach.

Question 2. (Optional) Was interesting or can we get a more elaborated statement $(2)$? Thanks in advance.

Now I am asking if $(2)$ is interesting versus Firoozbakht's conjecture or some unsolved problem related to the sum of divisor function, or well the theory of distribution of primes and the theory of the sum of divisors function itself. Or, if you can provide me a more elaborated statement inspired or similar than $(1)$ (I evoke that you can to combine Firoozbakht's conjecture with different arithmetic functions, like to primorials, the Euler's totient function or the sum of divisor function and different unsolved problems or theorems related to these arithmetic functions).

Answer 1

This is a simple proof by induction. The base case is $N=2$: $$\sigma(p_1p_2^2)=12\ge12=\frac32(p_3-1)\sigma(p_3)$$ Now, assume that for some $N$, we have: $$\sigma\left(\prod_{n=1}^{N}p_n^n\right)\ge\frac32(p_{N+1}-1)\prod_{n=2}^{N}p_{n+1}^{n-1}$$ this is the induction hypothesis. Now, for the induction step. Take the inequality you derived for $n=N+1$: $$\sigma(p_{N+1}^{N+1})>\frac{p_{N+2}-1}{p_{N+1}-1}\sigma(p_{N+2}^{N})$$ and multiply this by the inequality in the induction hypothesis to get: $$\sigma(p_{N+1}^{N+1})\sigma\left(\prod_{n=1}^{N}\sigma(p_n^n)\right)>\frac{3(p_{N+2}-1)}{2(p_{N+1}-1)}\cdot(p_{N+1}-1)\sigma(p_{N+2}^{N})\prod_{n=2}^{N}\sigma(p_{n+1}^{n-1})$$ Simplifying yields: $$\sigma\left(\prod_{n=1}^{N+1}p_n^n\right)>\frac32(p_{N+2}-1)\prod_{n=2}^{N+1}\sigma(p_{n+1}^{n-1})$$ And we're done. We've now proven your inequality, though it's not really strict (we have equality for $N=2$)