Trying to prove this relationship. Not sure if this is correct.

Question

I am trying to prove whether or not that:

The triangle inequality holds; that is, for every vectors $x$ and $y$, $$||x+y|| \leq ||x|| + ||y||,$$ for the vector function $$||x|| = \sum_{i=1}^n x^2_i.$$

I have this written currently and was wondering if it was sufficient to say that this inequality does not hold.

\begin{align} ||x+y|| &= \sum\limits_{i=1}^n (x_i+y_i)^2 \\ &= \sum_{i=1}^n (x^2_i + 2x_iy_i + y^2_i) \\ &= \sum\limits_{i=i}^n x^2_i + \sum\limits_{i=i}^n2x_iy_i + \sum\limits_{i=i}^ny^2_i \\ &=||x|| + ||y|| + \sum\limits_{i=i}^n2x_iy_i \\ &\neq ||x|| + ||y||. \end{align}

I’m a C.S. student so proofs aren’t really my strongsuit.

Answer 1

This inequality is false, picking up where you left off you know that $$||x+y|| = ||x|| + ||y|| + \sum_{i=1}^n 2x_iy_i$$ For the triangle inequality to be true you would need $\sum_{i=1}^n 2x_iy_i\leq 0$, for all $x_i,y_i$. But this is false, for example if $n=2$, let $x = (1,1)$ and $y = (1,1)$ we have $$\sum_{i=1}^n 2x_iy_i = 2(1*1 + 1*1) = 4>0$$ (Notice we have $||x+y|| = 2^2+2^2 = 8>||x||+||y|| = 2+2 = 4$)

Answer 2

Let $v$ be a vector with real scalars $a,b$ then $$\vert av+bv \vert = \vert(a+b)v \vert=\vert a +b \vert\vert v \vert=(a+b)^2\vert v\vert = \vert a^2v \vert + 2ab \vert v \vert + \vert b^2v \vert$$ and so we require $a = a^2$ and $b=b^2$, and $ab=0$. This gives $a(a-1)=0$ and likewise for $b$ so we see $a=b=0$ $a=1,b=0$ and $a=0,b=1$ are all solutions where equality holds.