Is $\sum\limits_{n=1} ^\infty |x_n||y_n| ≤\left(\sum\limits_{n=1} ^\infty |x_n|\right)\left(\sum\limits_{n=1} ^\infty |y_n|\right)$? Is $\sum\limits_{n=1} ^\infty |x_n||y_n| ≤ \left(\sum\limits_{n=1} ^\infty |x_n|\right)\left(\sum\limits_{n=1} ^\infty |y_n|\right)$?
I want to use this in a separate exercise but I can't think if this is true let alone how to prove it and I can't find any resources to help.
 A: Apply Cauchy-Schwarz
$$(x\cdot y)^2 \leq  ||x ||^2~||y||^2$$
to the vectors $x =(|x_1|,|x_2|,\ldots,|x_n|)$ and $y = (|y_1|,|y_2|,\ldots,|y_n|)$ to get
$$\left(\sum x_i y_i\right)^2 \leq \sum x_i^2\sum y_i^2$$
Now use that
$$\sum x_i^2 \sum y_i^2 \leq \left(\sum |x_i|\right)^2 \left(\sum |y_i|\right)^2$$ 
since $\left(\sum |x_i|\right)^2 = \sum x_i^2 + \sum_{i\not=j} |x_ix_j| \geq  \sum x_i^2$ and the result follows.
A: $$\sum\limits_{n=1} ^\infty |x_n||y_n|=|x_1||y_1|+\sum_{n=2}^\infty|x_n||y_n|\le |x_1|\sum_{n=1}^\infty|y_n|+\sum_{n=2}^\infty |x_n||y_n|\le$$
$$|x_1|\sum_{n=1}^\infty|y_n|+|x_2||y_2|+\sum_{n=3}^\infty |x_n||y_n|\le |x_1|\sum_{n=1}^\infty|y_n|+|x_2|\sum_{n=1}^\infty|y_n|+\sum_{n=3}^\infty |x_n||y_n|\le$$
$$\vdots$$
$$|x_1|\sum_{n=1}^\infty|y_n|+|x_2|\sum_{n=1}^\infty|y_n|+|x_3|\sum_{n=1}^\infty|y_n|+\cdots =$$
$$ \sum\limits_{n=1} ^\infty |x_n|\sum\limits_{n=1} ^\infty |y_n|$$
A: How about something more elementary?
If $x_i=0$ for all $i$, it is easy.  So assume not.
Let $M = \max\{|y_1|,|y_2|,\dots\}$.  If $M=\infty$ it is easy, so assume not.
Certainly $\sum_{i=1}^\infty |y_i| \ge M$.
For all $i$, we have $|x_i y_i| \le |x_i|\; M$, so
$$
\sum_{i=1}^\infty |x_i y_i| \le \sum_{i=1}^\infty \big(|x_i|M\big) =
\left(\sum_{i=1}^\infty |x_i|\right)M \le
\sum_{i=1}^\infty |x_i| \sum_{i=1}^\infty |y_i|
$$
