Simple inequality with products of finite geometric series Given numbers $p$ and $q$ with $1\le p < q$, and integers $n$ and $m$ with $1\le n < m$, does this simple inequality hold:
$$\left(1+\frac{1}{p}+\frac{1}{p^2}+\frac{1}{p^3}+\dots+\frac{1}{p^n}\right)\left(1+\frac{1}{q}+\frac{1}{q^2}+\frac{1}{q^3}+\dots+\frac{1}{q^m}\right) < \left(1+\frac{1}{p}+\frac{1}{p^2}+\frac{1}{p^3}+\dots+\frac{1}{p^m}\right)\left(1+\frac{1}{q}+\frac{1}{q^2}+\frac{1}{q^3}+\dots+\frac{1}{q^n}\right)$$
Must be elementary. (Edit: I am happy with any proof or disproof, I just meant to say I thought the inequality in itself should be "elementary".)

When $p$ and $q$ are prime numbers, this says $\sigma_{-1}(p^nq^m)<\sigma_{-1}(p^mq^n)$ where $\sigma_{-1}$ is the abundancy index.
 A: Yes, it holds.


*

*If $1=p\lt q$, then the inequality is equivalent to$$(n+1)\cdot\frac{1-\frac{1}{q^{m+1}}}{1-\frac 1q}\lt (m+1)\cdot\frac{1-\frac{1}{q^{n+1}}}{1-\frac 1q},$$i.e.$$\frac{1-\frac{1}{q^{m+1}}}{1+m}< \frac{1-\frac{1}{q^{n+1}}}{1+n}\tag1$$Let $f(x):=\frac{1-\frac{1}{q^{x+1}}}{1+x}$ for $x\ge 1$. Then, $$f'(x)=\frac{-q^{x + 1} + x \ln q + \ln q + 1}{q^{x+1}(x + 1)^2}=\frac{g(x)}{q^{x+1}(x+1)^2}$$where $g(x):=-q^{x + 1} + x \ln q + \ln q + 1$. Then,$$g'(x)=-(q^{x+1}-1)\ln q\lt 0$$So, $g(x)$ is decreasing with $g(1)=0$ implying $g(x)\lt 0$ for $1\lt x$. Since $f'(x)\lt 0$, we have that  $f(x)$ is decreasing, from which $(1)$ follows. 

*If $1\lt p\lt q$, then the inequality is equivalent to$$\frac{1-\frac{1}{p^{n+1}}}{1-\frac 1p}\cdot\frac{1-\frac{1}{q^{m+1}}}{1-\frac 1q}\lt \frac{1-\frac{1}{p^{m+1}}}{1-\frac 1p}\cdot \frac{1-\frac{1}{q^{n+1}}}{1-\frac 1q},$$i.e.$$\left(1-\frac{1}{p^{n+1}}\right)\left(1-\frac{1}{q^{m+1}}\right)\lt \left(1-\frac{1}{p^{m+1}}\right)\left(1-\frac{1}{q^{n+1}}\right),$$i.e.$$\frac{1-\frac{1}{q^{m+1}}}{1-\frac{1}{p^{m+1}}}\lt \frac{1-\frac{1}{q^{n+1}}}{1-\frac{1}{p^{n+1}}},$$i.e. $$\frac{p^{m+1}(q^{m+1}-1)}{q^{m+1}(p^{m+1}-1)}\lt \frac{p^{n+1}(q^{n+1}-1)}{q^{n+1}(p^{n+1}-1)}\tag2$$Let $h(x):=\frac{p^{x+1}(q^{x+1}-1)}{q^{x+1}(p^{x+1}-1)}$ for $x\ge 1$. Then, we get$$h'(x)=\frac{(\frac pq)^{x+1}(p^{x+1}-1)(q^{x+1}-1)\left(\frac{\ln q}{q^{x+1}-1}-\frac{\ln p}{p^{x+1}-1}\right)}{(p^{x+1}-1)^2}$$Let $i(x):=\frac{\ln x}{x^{c+1}-1}$ where $c\gt 1$ is a real number. Then,
$$i'(x)=-\frac{x^{c+1}((c+1)\ln x-1)+1}{x(c^{x+1}-1)^2}$$For $x\ge 2$, we get $(c+1)\ln x-1\gt (1+1)\ln 2-1\gt 0$ from which $i'(x)\lt 0$ follows. So, for $x\ge 2$, we see that $i(x)$ is decreasing from which $\frac{\ln q}{q^{x+1}-1}-\frac{\ln p}{p^{x+1}-1}\lt 0$. It follows that $h'(x)\lt 0$, so $h(x)$ is decreasing for $x\ge 2$. Now, we can see that $h(1)\gt h(2)$ since this is equivalent to$$\frac{p^{2}(q^{2}-1)}{q^{2}(p^{2}-1)}\gt \frac{p^{3}(q^{3}-1)}{q^{3}(p^{3}-1)},$$i.e.$$q(q+1)(p^2+p+1)\gt p(p+1)(q^2+q+1),$$i.e.$$(q-p) (p + q + 1)\gt 0,$$i.e.$$q\gt p$$Therefore, we see that $h(1)\gt h(2)\gt h(3)\gt\cdots$. It follows that $(2)$ is true for integers $n,m$ such that $1\le n\lt m$.
