General Proof for the triangle inequality

As @ivan indicates, the inequality is reversed - it should be

$$ |x_1 + x_2 + \dots + x_n| \leq |x_1| + |x_2| + \dots + |x_n| $$

As the base case for induction, you need to show (or assert? can you take the "basic" triangle inequality for granted?)

$$ |x_1 + x_2| \leq |x_1| + |x_2|. $$

Hint:

One way to do this is to show $(|a + b|)^2 \leq (|a| + |b|)^2$ by expanding the LHS and using $ab \leq |a||b|$.

Then, for induction, assume

$$ |x_1 + x_2 + \dots + x_n| \leq |x_1| + |x_2| + \dots + |x_n| $$

and show

$$ |x_1 + x_2 + \dots + x_n + x_{n+1}| \leq |x_1| + |x_2| + \dots + |x_n| + |x_{n+1}|$$

using the induction hypothesis and the base case.

It is ok for $|x_1 + x_2| \le |x_1| + |x_2|$ (I)

$|x_1 + x_2 +\cdots+ x_k| \le |x_1| + |x_2|+\cdots+ |x_k|$ (Hypothesis)(II)

For $n = k+1$ :

a) 1st. Apply The triangle inequality for $2$ different "numbers" $(x_1 + x_2 +\cdots+ x_k)$ and $x_{k+1}$ because it is ok by (I)

b)2nd Apply the induction Hypothesis (II)

$|x_1+x_2+\cdots+ x_k +x_{k+1}| \le |x_1 + x_2 +\cdots+x_k|+|x_{k+1}| \le |x_1|+|x_2|+\cdots+|x_k|+|x_{k+1}|$