1. Triangle inequality proof

Q: Prove that $\displaystyle ||\vec{v}+\vec{u}||=||\vec{u}||+||\vec{v}||$ if and only if $\displaystyle \vec{u}$ and $\displaystyle \vec{v}$ have the same direction.

A: Well, if $\displaystyle \vec{u}$ the same direction of $\displaystyle \vec{v}$, then $\displaystyle \vec{u}$ must be a multiple of $\displaystyle \vec{v}$. Thus, $\displaystyle \vec{u}=k\vec{v}$ where $\displaystyle k\in{\mathbb{R}}$ (I am not sure if I lost any generality by choosing $\displaystyle \vec{u}$). From here I expanded $\displaystyle ||\vec{v}+\vec{u}||$ and ended up with some messy algebra and not the result I was looking for.

In the other direction I started with $\displaystyle ||\vec{v}+\vec{u}||=...$ and worked my way down to $\displaystyle =||\vec{u}||^{2}+|2<\vec{u},\vec{v}>|+||\vec{v}||^ {2}$ and figured I have to introduce the Cauchy-Schawrz Ineguality here, but I am not sure how.

I am having a hard time formalizing this proof. All I have is a bunch of dead und scratch work. It seems that this proof should flow with ease, because, geometrically, it is pretty obviouse.

I would appreciate some guidence.

Thank you.

2. Let $\displaystyle \vec{u} = \left<u_1, u_2, \cdots, u_n\right>$ and $\displaystyle \vec{v} = \left<v_1, v_2, \cdots, v_n\right>$.

As you noted, $\displaystyle \vec{u} = k\vec{v}$ and so:

\displaystyle \begin{aligned} \left| \vec{u} + \vec{v}\right| & = \sqrt{\left(kv_1 + v_1\right)^2 + \left(kv_2 + v_2\right)^2 + \cdots + \left(kv_n + v_n\right)^2} \\ & = \sqrt{ \left[(k+1)v_1\right]^2 + \left[(k+1)v_2\right]^2 + \cdots + \left[(k+1)v_n\right]^2 } \\ & \ \vdots \\ & = k|v| + |v| \end{aligned}

3. Hello,
Originally Posted by o_O
Let $\displaystyle \vec{u} = \left<u_1, u_2, \cdots, u_n\right>$ and $\displaystyle \vec{v} = \left<v_1, v_2, \cdots, v_n\right>$.

As you noted, $\displaystyle \vec{u} = k\vec{v}$ and so:

\displaystyle \begin{aligned} \left| \vec{u} + \vec{v}\right| & = \sqrt{\left(kv_1 + v_1\right)^2 + \left(kv_2 + v_2\right)^2 + \cdots + \left(kv_n + v_n\right)^2} \\ & = \sqrt{ \left[(k+1)v_1\right]^2 + \left[(k+1)v_2\right]^2 + \cdots + \left[(k+1)v_n\right]^2 } \\ & \ \vdots \\ & = k|v| + |v| \end{aligned}
Are you sure you can take wlog this norm ? Though all norms are equivalent in finite dimensions, we're talking about a norm in general, here.
What you didn't say, and that is helpful for the solution, is that k has to be positive.

Otherwise,
Originally Posted by Danneedshelp
Q: Prove that $\displaystyle ||\vec{v}+\vec{u}||=||\vec{u}||+||\vec{v}||$ if and only if $\displaystyle \vec{u}$ and $\displaystyle \vec{v}$ have the same direction.
They wouldn't have the same direction as they would go in opposit directions.

A: Well, if $\displaystyle \vec{u}$ the same direction of $\displaystyle \vec{v}$, then $\displaystyle \vec{u}$ must be a multiple of $\displaystyle \vec{v}$. Thus, $\displaystyle \vec{u}=k\vec{v}$ where $\displaystyle k\in{\mathbb{R}}$ (I am not sure if I lost any generality by choosing $\displaystyle \vec{u}$).
It's okay, there is no loss of generality. You choose u because it's the assumption of the implication.

From here I expanded $\displaystyle ||\vec{v}+\vec{u}||$ and ended up with some messy algebra and not the result I was looking for.
Actually, you just have to use the axioms/definition of a norm.

There is the first axiom (positive homogeneity) : $\displaystyle \|a \vec{u}\|=|a| ~\|\vec{u}\|$, where a is a real number.

So in this case (I'll forget the \vec...too long ^^), $\displaystyle \|u+v\|=\|kv+v\|=(k+1)~ \|v\|=k \|v\|+\|v\|=\|kv\|+\|v\|=\|u\|+\|v\|$

In the other direction I started with $\displaystyle ||\vec{v}+\vec{u}||=...$ and worked my way down to $\displaystyle =||\vec{u}||^{2}+|2<\vec{u},\vec{v}>|+||\vec{v}||^ {2}$ and figured I have to introduce the Cauchy-Schawrz Ineguality here, but I am not sure how.

I am having a hard time formalizing this proof. All I have is a bunch of dead und scratch work. It seems that this proof should flow with ease, because, geometrically, it is pretty obviouse.

I would appreciate some guidence.

Thank you.
A scalar product gives a norm.
So let's define <,> as the scalar product associated with this norm.

Now, we assume that $\displaystyle \|u+v\|=\|u\|+\|v\|$ and we want to prove that there exists $\displaystyle k\in\mathbb{R}_+$ such that $\displaystyle u=kv$

$\displaystyle \|u+v\|^2=\langle u+v,u+v\rangle =\dots=\|u\|^2+\|v\|^2+2\langle u,v\rangle$ by definition.

Now, we know that $\displaystyle \|u+v\|=\|u\|+\|v\| \Rightarrow \|u+v\|^2=\|u\|^2+\|v\|^2+2\|u\|\|v\|$

So $\displaystyle \langle u,v\rangle=\|u\|\|v\|$

And here, apply Cauchy-Schwarz inequality, stating that there is equality if and only if u and v are linearly dependent.
That is to say u=kv, but where k is not necessarily positive.

However, the equivalence only stands for k positive.

4. Here's a little trick that avoids using Cauchy–Schwarz. You have already seen that if $\displaystyle \|u+v\| = \|u\|+\|v\|$ then $\displaystyle \langle u,v\rangle = \|u\|\|v\|$. Let $\displaystyle w = \|v\|u-\|u\|v$. Then $\displaystyle \|w\|^2 = \langle w,w\rangle = \langle \|v\|u-\|u\|v,\|v\|u-\|u\|v\rangle$, which simplifies to $\displaystyle 2\|u\|v\|(\|u\|\|v\| - \langle u,v\rangle)$. If $\displaystyle \langle u,v\rangle = \|u\|\|v\|$ then it follows that $\displaystyle w=0$, from which $\displaystyle v = \tfrac{\|v\|}{\|u\|}u$.