Absolute value inequality: $ |x_n y_n-x_n y|≤|x_n ||y_n-y|$

Question

I am trying the prove the following theorem:

If $(x_n)$,$(y_n)$ are sequences with $x_n→x$,$y_n→y$ and $λ∈R$.

Then: $x_n y_n→xy$

A section fo my proof:

$$|x_n y_n-xy| =|x_n y_n-x_n y+x_n y-xy|$$ $$ ≤|x_n y_n-x_n y|+|x_n y-xy|$$ $$≤|x_n ||y_n-y|+(|y|+1)|x_n-x|$$

I know the general sense to prove above theorem, however i am confused at the absolute value inequality.

For (1) and (2), which of the inequlaity/equality is correct.

1. $ |x_n y_n-x_n y|≤|x_n ||y_n-y|$ or $ |x_n y_n-x_n y|=|x_n ||y_n-y|$
2. $ |x_n y-xy|≤||y||x_n-x|$ or $ |x_n y-xy|=|y||x_n-x|$

Answer 1

$$|x_ny_n - x_ny| = |x_n(y_n-y)| = |x_n||y_n-y|\leq |x_n||y_n-n|$$ so for (1), both the inequality and the equality is true.

Similarly, $$|x_ny-xy| = |y(x_n-x)|=|y||x_n-x| \leq |y||x_n-x|$$

so both are also true for (2).

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So, sure, in the proof, they could also use the inequality, but it makes sense sometimes to use the inequality, especially if it is in a string of other inequalities.

What I mean by that is that if I want to prove that $a\leq h$, sometimes, it's better to write out the proof as

$$a\leq b\leq c\leq d\leq e\leq f\leq g\leq h$$

even if it would strictly speaking be true that

$$a\leq b=c=d=e=f=g=h.$$

In that second proof, it might confuse a sloppy reader to think an equality was proven.

Answer 2

The absolute value is multiplicative, i.e. for every $a,b \in \mathbb{R}$ we have $$|a\cdot b| = |a|\cdot |b|$$ (you can easily show this by case distinction on the signs of $a$ and $b$).

So in both (1) and (2) equality holds.