Prove that in $\mathbb{R}^{4}$ that the $1$-norm satisfies the triangle inequality, given that for real numbers $\lVert x+y\rVert ≤ |x| + |y|$

Question

I tried using an example where I added the norm of $2$ different vectors together vs I added 2 vectors first and found its norm. However, the result only turned out that $|x+y|=|x| + |y|$. How can I prove the "less than" part of the inequality?

Answer 1

It is not true that $|x+y|=|x|+|y|$ for real numbers $x$ and $y$. Take $x=1,y=-1$ for a counter-example. What is true is $|x+y| \leq |x|+|y|$.

Using this we get $$\|(a,b,c,d)+(a',b',c',d')\|=\|(a+a',b+b',c+c',d+d')\|$$ $$=|a+a'|+|b+b'|+|c+c'|+|d+d'| \leq (|a|+|a'|)+ (|b|+|b'|)+ (|c|+|c'|)+ (|d|+|d'|)=\|(a,b,c,d)\|+\|(a',b',c',d')\|$$