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Suppose we have 6 sequences $a_n,b_n,c_n$ and $p_n,q_n,r_n$, all of which are positive (i.e. $a_n,b_n,c_n,p_n,q_n,r_n \geq 0, \forall n\in\mathbb{N}$).

Does the following implication hold? \begin{align}\left(\sum\limits_{n=1}^{\infty} a_nb_n > \sum\limits_{n=1}^{\infty} p_nq_n\right) &\wedge \left(\sum\limits_{n=1}^{\infty} a_nc_n > \sum\limits_{n=1}^{\infty} p_nr_n\right) \\\Rightarrow\sum\limits_{n=1}^{\infty} a_nb_nc_n &> \sum\limits_{n=1}^{\infty} p_nq_nr_n\end{align}

We assume that all of the sums are convergent.

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No.

Let $a_n=\frac{1}{n^5}$, $b_n=c_n=\frac{1}{n^4}$, $p_n=\frac{1}{n^8}$, $r_n=q_n=\frac{1}{n^2}$. Then

$\sum_{n=1}^{\infty} a_n b_n=\sum_{n=1}^{\infty} a_n c_n=\sum_{n=1}^{\infty}\frac{1}{n^9}>\sum_{n=1}^{\infty}\frac{1}{n^{10}}=\sum_{n=1}^{\infty} p_n q_n=\sum p_n r_n$

but

$\sum_{n=1}^{\infty} a_n b_n c_n = \sum_{n=1}^{\infty}\frac{1}{n^{13}} < \sum_{n=1}^{\infty}\frac{1}{n^{12}} = \sum_{n=1}^{\infty} p_n q_n r_n$

I am sure there are simpler examples but that is the one that came to mind. Technically of course I'd need to prove that $\sum_{n=1}^{\infty}\frac{1}{n^k}>\sum_{n=1}^{\infty}\frac{1}{n^{k'}}$ for $2\leq k

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    Yeah, maybe there is a simpler example, but this is enough to convince me. Thanks. – 2017-01-26