# Homework Help: Triangle Inequality for integrals proof

1. Oct 16, 2011

### Henry365

1. The problem statement, all variables and given/known data
What I want to show is this:
∫|x+y| ≤ ∫|x| + ∫|y|

2. Relevant equations
|x+y| ≤ |x| + |y|

3. 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.

2. May 13, 2013

### 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
$$\int_{a}^{b}f(x) \, dx \le \int_{a}^{b} g(x) \, dx.$$

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