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Prove the following inequality

$$\frac{H_p}{p}\prod_{m=2}^{p}\sum_{n=0}^{\infty}\frac{1}{(n+1)(mn+1)}\ge 1$$

for positive integer $p$ and $H_p=\sum_{l=1}^{p}\frac{1}{l}$.

In fact the origin of this problem is the following integral inequality :

$$H_n \left(\int_{0}^{1} \frac{x-1}{x^2-1}dx\right)\left(\int_{0}^{1} \frac{x^2-1}{x^3-1}dx\right)\cdots\left(\int_{0}^{1} \frac{x^{n-1}-1}{x^n-1}dx\right)\ge 1 $$

when I developed this inequality to obtain its discrete formulation I got the above inequality.

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    I'm not sure this is true as stated? The sum is approximately $\sum_n\frac{1}{mn^2}$ which converges to $\frac{\pi^2}{6m}$, so the overall term is about $\frac{H_p}{p\cdot p!}\left(\frac{\pi^2}{6}\right)^p \to 0$ – 2017-01-06
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    We don't like demands. We also like when people show us what they know or have tried. Please edit your question to include that information. – 2017-01-06
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    @ChrisJones The first term in the sum is always $1$, so it cannot behave asymptotically like $\frac{\pi^2}{6m}$. In general it is hard to estimate the size of a convergent sum by looking only at the asymptotics of the tail (large $n$), since much of the mass is in the small values of $n$. – 2017-01-07

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This is not an answer but, may be, it could help.

$$\int \frac{x^{n-1}-1}{x^n-1}\,dx=\frac{n x^{n+1} \, _2F_1\left(1,1+\frac{1}{n};2+\frac{1}{n};x^n\right)+(n+1) \left(\log \left(1-x^n\right)+n x\right)}{n (n+1)}$$ $$\int_{0}^{1} \frac{x^{n-1}-1}{x^n-1}\,dx=1-\frac{H_{\frac{1}{n}}}{n} \qquad \qquad(\Re(n)>0)$$ which make the expression to be $$A_n=H_n\prod _{i=2}^n \left(1-\frac{H_{\frac{1}{i}}}{i}\right)$$ So $$\frac{A_{n+1}}{A_n}=\left(1-\frac{H_{\frac{1}{n+1}}}{n+1}\right)\frac{ H_{n+1}}{H_n}$$ Now, for large values of $n$, the asymptotics is $$\frac{A_{n+1}}{A_n}=1+\frac{1}{n \left(\gamma +\log \left({n}\right)\right)}+O\left(\frac{1}{n^2}\right)\implies {A_{n+1}} >{A_n}$$ and since $A_1=1$ the inequality holds.

Just to confirm, I produced below a table of $A_n$ $$\left( \begin{array}{cc} n & A_n \\ 1 & 1.00000 \\ 2 & 1.03972 \\ 3 & 1.08220 \\ 4 & 1.12224 \\ 5 & 1.15908 \\ 6 & 1.19288 \\ 7 & 1.22398 \\ 8 & 1.25272 \\ 9 & 1.27941 \\ 10 & 1.30431 \end{array} \right)$$