Prove that $|x+y+z| \le |x|+|y|+|z|$

Question

Prove the following: $$|x+y+z| \le |x|+|y|+|z|$$ It is so trivial that I do not have idea how to show it. Thus, how do I show it?

Answer 1

If you are asked to prove the triangle inequality, your proof may be longer. But if you can just invoke the triangle inequality, then you have $$|x+y+z| \le |x+y|+|z| \le |x|+|y|+|z|$$ or $$|x+y+z| \le |x|+|y+z| \le |x|+|y|+|z|.$$ or $$|x+y+z| \le |y|+|x+z| \le |x|+|y|+|z|.$$

Answer 2

Let $x, y, z$ be vectors in the Euclidean space. Then your inequality immediately follows from the well-known fact that a straight line is the shortest path between 2 points. If $x, y, z$ are just real numbers, it is just a special case where all 3 vectors are collinear.

Answer 3

If you are working with real numbers, we know that $|x|\ge 0,$ $|x|\ge x$ and $|x|^2=x^2.$ $$|x|-x\ge 0\iff (1+2|x|+|x|^2)-(1+2x+x^2)=(1+|x|)^2-|1+x|^2\ge0 .$$ This implies $$1+|x|\ge|1+x|,\,\,\,\,\,\forall x\in\mathbb{R}. $$ Then substituting $x=\dfrac{a}{b}$ you can obtain the triangle inequality $$|a|+|b|\ge|a+b|,\,\,\,\,\,\forall a,b\in\mathbb{R}.$$ Apply this inequality two times!!
Good Luck.