How can I formally prove that
$$\max\{\lvert x+y\rvert _i \} \leq \max\{ \lvert x_j \rvert \} + \max\{\lvert y_k \rvert \}$$
Where $x,y$ are the components of a $n$-vector with $1 \leq i,j,k \leq n$
It's obvious that if either $x$ or $y$ on the left side are negative the inequality is fulfilled with $\lt $ . But how do I prove that formally? Is a simple case analysis enough?
Let $m=\max\{|(x+y)_i|\}$. Then the maximium is achieved for some index, $\nu$ say. Then $$m=|(x+y)_\nu|\stackrel{(1)}=|x_\nu+y_\nu|\stackrel{(2)}\le |x_\nu|+|y_\nu|\stackrel{(3)}\le \max\{|x_j|\}+\max\{|y_k|\}$$ where $(1)$ follows from the definition of vector addition, $(2)$ is the triangle inequality, and $(3)$ follows from the definition of maximum.
Here is a formal proof. Fix $1\leq i \leq n$, then it is always the case that $$ |x_i|\leq \max\{|x_j|\}. $$ In particular, and using the triangle inequality, we have $$ |x_i+y_i|\leq |x_i|+|y_i|\leq \max\{|x_j|\} + \max\{|y_k|\}. $$ Since we pick an arbitrary $i$, the previous inequality is true for all $1\leq i \leq n$, hence also true for the $i$ giving the maximum. In other words, $$ \max\{|x_i+y_i|\}\leq \max\{|x_j|\} + \max\{|y_k|\}. $$
For individual components, you have $|x+y|_i\le |x_i|+|y_i|, 1\le i\le n$, thus, from the definition $|x+y|_i\le \max_j|x_j|+\max_k|y_k|\forall i\implies \max_i|x+y|_i\le \max_j|x_j|+\max_k|y_k|$
max{∣x+y∣i} ≤ max{∣x∣i} + max{∣y∣i} by the triangle inequality
clearly,
max{∣x∣i} ≤ max{∣x∣j} (think about it)
max{∣y∣i} ≤ max{∣y∣k}
Therefore,
max{∣x+y∣i} ≤ max{∣x∣j} + max{∣y∣k}
