Triangle Inequality for integrals proof

[Henry365]

Homework Statement

What I want to show is this:
∫|x+y| ≤ ∫|x| + ∫|y|

Homework Equations

|x+y| ≤ |x| + |y|

The Attempt at a Solution

So I thought if I used the triangle inequality I could get to something along the lines of:

Lets g belong to the real numbers
∫|x+y| = ∫|x+g-g+y|≤ ∫|x+g| + |y-g|= ∫|x+g| + ∫|y-g|

As g belongs to the reals it can be zero meaning ∫|x+y| ≤ ∫|x| + ∫|y|.

Now the problem with this is that is uses the triangle inequality and I have no idea if the triangle inequality works this way, and if it does I need to prove it, and I have no idea about where to start that from.

 

[Ackbach]

I suppose I am necro-posting here, but this result follows from two facts. One is the linearity of integration:
\int (f(x)+g(x)) \, dx = \int f(x) \, dx + \int g(x) \, dx,
and the other is that integration preserves inequalities: if f(x) \le g(x) on the interval [a,b], then
\int_{a}^{b}f(x) \, dx \le \int_{a}^{b} g(x) \, dx.