I managed to prove the triangle inequality with linear algebra however I feel like there is some algebraic identity that trivializes this inequality.
Let $a=[a_1, a_2, ... a_n],b=[b_1, b_2, ... b_n]\ \epsilon\ \mathbb{R}^n$, then we have $$\sqrt{a_1^2+a_2^2...a_n^2}+\sqrt{a_1^2+a_2^2...a_n^2}\geq \sqrt{(a_1+b_1)^2+(a_2+b_2)^2...(a_n+b_n)^2}$$.
My solution is to simply square both sides and the rest can be simplified by cancellation and CS inequality. However, it looks like there is an identity (maybe Holder's?) that directly trivializes this problem. Is this true?
If you use Cauchy-Schwarz, \begin{align} \|a+b\|^2&=(a+b)^T(a+b)=a^Ta+b^Tb+2b^Ta\\[0.2cm] &\leq a^Ta+b^Tb+2\|a\|\,\|b\|\\[0.2cm] &=\|a\|^2+\|b\|^2+2\|a\|\,\|b\|=(\|a\|+\|b\|)^2. \end{align} Taking square roots (of positive numbers), you get $$ \|a+b\|\leq\|a\|+\|b\|. $$