Prove that for every $n \in \mathbb{N}$ and for all real numbers $x_1,x_2....x_n \in \mathbb{R}$ Prove that for every $n \in \mathbb{N}$ and for all real numbers $x_1,x_2....x_n \in \mathbb{R}$,
$$|x_1 + x_2 +....+x_n| \leq |x_1| + |x_2| +...+|x_n|$$
My attempt:
Base case: $n = 1$
$$|x_1| \leq |x_1|$$
Induction step: $n + 1$
$$|x_1 + x_{n+1}| \leq |x_1| + |x_{n+1}|$$
By the triangle inequality we have shown the inequality to be true for $n + 1$
Is this a valid and good proof? Is there something I am missing or should add?
 A: Induction step, If the statement holds for $n$, then it is to be shown that it holds for $n+1$. That is, the following should be shown to be true
$$
|x_1+\dots x_n| \leq |x_1|+\dots |x_n| \overset{by\ triangle\ inequality}{\Longrightarrow} |x_1+\dots +x_{n+1}| \leq |x_1|+\dots |x_{n+1}|
$$
That's how it goes
$$
(\text{by triangle inequality})\ |x_1+\dots +x_n+x_{n+1}| \leq |x_1+\dots+x_n|+|x_{n+1}| \\
(\text{by the assumption for } n)\Rightarrow |x_1+\dots +x_{n+1}| \leq |x_1|+\dots +|x_n|+|x_{n+1}|
$$
A: hint
For $ n=1$, it is obvious.
For $ n=2 $, we prove by disjunction of cases that
$$|x_1+x_2|\le |x_1|+|x_2|$$
for the $(n+1)^{\text{th}}$ step, you have to check that
$$|x_1+x_2+...+x_n+x_{n+1}|\le$$
$$|x_1|+|x_2|+...|x_n|+|x_{n+1}|$$
A: To do a proof by induction you need to assume that the statement is true for $n$ and then show it holds for $n + 1$. One way is the following:
Let's assume that $|x_1 + ... + x_n| \leq |x_1|+ ... + |x_n|$
Then $|\underbrace{x_1 + ... + x_n}_k + x_{n + 1}| \leq |k| + |x_{n + 1}|$. This is just the triangle inequality. But $|k |\leq |x_1|+ ... + |x_n|$. So this means that:
$$|\underbrace{x_1 + ... + x_n}_k + x_{n + 1}| \leq |k| + |x_{n + 1}|\leq |x_1|+ ... + |x_n| + |x_{n + 1}|$$
You know it's true for $n = 2$ and here you have it
A: To show the inductive step I would use your idea of triangle inequality:
$$
|x_1+\dots +x_n| \leq |x_1|+\dots |x_n| \implies |x_1+\dots+ x_n| +|x_{n+1}| \leq |x_1|+\dots +|x_n|+|x_{n+1}| $$
but by triangle inequality $ |x_1+\dots x_n+x_{n+1}| \leq|x_1+\dots x_n|+|x_{n+1}|$.
