Sum of Norms vs Norm of Sums Proof I know how to prove the triangle inequality. How can we prove that $$\left|\left|\sum_{i=1}^nx_i\right|\right|\leq \sum_{i=1}^n \left|\left|x_i\right|\right|,$$
where $x_i$ are vectors in $R^k$.
Does the proof somehow follows from the triangle inequality?
 A: We prove the claim is true for every $n\in \mathbb Z^+$ by induction.
For $n=1$ it is clear and for $n=2$ it is the normal triangle inequality. Suppose it has been proved for $n$.
We then have:
$||\sum\limits_{i=1}^{n+1}x_i||=||\sum\limits_{i=1}^{n}x_i+x_{n+1}||\leq ||\sum\limits_{i=1}^{n}x_i||+||x_{n+1}||\leq \sum\limits_{i=1}^{n}||x_i||+||x_{n+1}||=\sum\limits_{i=1}^{n+1}||x_i||$
A: The base of induction is obvious.
Let $$\sum_{k=1}^n\|x_i\|\geq\|\sum_{k=1}^nx_i\|.$$
Thus, by the triangle inequality
$$\sum_{k=1}^{n+1}\|x_i\|=\sum_{k=1}^n\|x_i\|+\|x_{n+1}\|\geq$$
$$\geq \|\sum_{k=1}^nx_i\|+\|x_{n+1}\|\geq\|\sum_{k=1}^{n+1}x_i\|$$
and we are done!
A: If you have already proven triangle inequality is true the you must have $$||x_1+x_2|| \le ||x_1||+||x_2||$$
This is the base case of induction. Then assume that this is true for $n$ terms. 
$$||\sum_1^nx_i|| \le \sum_1^n||x_i||$$
Then you must show that 
$$||\sum_1^{n+1}x_i||\le\sum_1^{n+1}||x_i||$$. 
This can be shown as, since first and second relations are true $$||\sum_1^{n+1}x_i||\le||\sum_1^nx_i||+||x_{n+1}||\le\sum_1^n||x_i||+||x_{n+1}||=\sum_1^{n+1}||x_i||$$ 
