# Inequality on lengths and sums of vectors $\left\lVert\sum_i \vec{a_i}\right\rVert \le \sum_i \left\lVert \vec{a_i}\right\rVert$

I'm trying to show the following inequality, which expresses the fact that the magnitude of the sum of some vectors, is less than the sum of the individual magnitudes: $$\left\lVert\sum_i \vec{a_i}\right\rVert \le \sum_i \left\lVert \vec{a_i}\right\rVert$$ I found this on a page with Cauchy Schwarz problems, but stated without proof.

I've also be considered if it can be solved by induction, but I haven't got an idea how.

• This sum has finite number of terms? – user73454 May 10 '14 at 15:27
• Yes, sorry for not making that clear. – Thomas Ahle May 10 '14 at 16:06

## 1 Answer

Triangle Inequality: $\|v+w\|\leq \|v\|+\|w\|$. Thus, $$\|\sum\limits_{i=1}^n v_i\|=\|\sum\limits_{i=1}^{n-1} v_i+v_n\|\leq\| \sum\limits_{i=1}^{n-1} v_i\|+\|v_n\|,$$ then applies induction, i.e., $\| \sum\limits_{i=1}^{n-1} v_i\|\leq \sum\limits_{i=1}^{n-1} \|v_i\|$.

• Doh, you are right. I guess I was just specifically looking for a Cauchy Schwarz proof. – Thomas Ahle May 10 '14 at 15:49
• @ThomasAhle, Let $t=sign(\langle v, w\rangle)$, then $$\| v+tw\|^2=\| v\|^2+2t\langle v, w\rangle +\|w\|^2,$$ but $\|v+tw\|\let \|v\|+\|w\|$, then $$\|v+tw\|^2\leq\|v\|^2+2\|v\|\|w\| +\|w\|^2.$$ Thus $$t\langle v,w\rangle\leq \|v\|\|w\|$$ i.e., $$|\langle v,w\rangle|\leq \| v\|\|w\|.$$ – user73454 May 10 '14 at 16:30
• @ThomasAhle, there is problem on to compile, but let $t=sign(\langle v,w\rangle)$, then $$|v+tw|^2=|v|^2+2t\langle v,w\rangle+|w|^2$$ but $|v+tw|\leq |v|+|t||w|\leq |v|+|w|$, then $$|v + tw|^2\leq |v|^2+2|v||w|+|w|^2.$$ Thus $$t\langle v,w\rangle=|\langle v,w\rangle|\leq |v||w|.$$ – user73454 Dec 29 '14 at 18:20