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?

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
