Is there a nicer way to find this number? I was solving the following problem: Let $x,y,z \in \mathbb{R}^n$ be such that using euclidean norm $|x-z|=|x-y|+|y-z|$, prove that there is $t \in [0,1]$ such that $y=(1-t)x+tz$.
Well, I first noticed that $y=(1-t)x+tz$ is equivalent to have $(y-x)=t(z-x)$. Now, I've proved already that in the euclidean norm if $|u+v|=|u|+|v|$ then there is $\alpha \geq 0$ such that $v = \alpha u$. So, I've set $u =x-y$ and $v=y-z$ and use this fact to deduce that $y-z=\alpha(x-y)$. I've tried a lot to make this look like what I wanted but it wasn't possible.
So I've changed roles, I've set $u = y-z$ and $v =x-y$, then I found $x-y = \alpha(y-z)$ and so I've had the intuition to sum and subtract $x$ to make the term $z-x$ appear, getting $x-y = \alpha(y-x+x-z)$, this lead my quickly to $(y-x)=\alpha/(1+\alpha)(z-x)$. Since $t=\alpha/(1+\alpha) \in [0,1]$, this shows that such $\alpha$ exists.
The only thing making me a little confused is that I needed to take a lot of guesses and tries along this proof. It wasn't a straightforward way to find this $t$. Is there a nicer way to solve this or the way I've thought is the right way in this kind of problem: we simply take guesses trying to make appear terms we know how to relate, try those guesses and if they fail try again?
Thanks very much in advance!
 A: I think a tiny amount of looking at what your statement means geometrically helps here.
The question asks you to prove that if $|x-z| = |x-y| + |y-z|$, then $y = (1-t)x + tz$ for some $t \in [0, 1]$. This is the algebraic statement.  Geometrically, what this means is that if the distance from $x$ to $z$ is the sum of distances from $x$ to $y$ and from $y$ to $z$, then $y$ is on the line (segment) joining $x$ to $z$. (The geometric meaning of $y = (1-t)x + tz$ is that $y$ is on the line segment joining $x$ to $z$.)
You've already shown that if $|u + v| = |u| + |v|$, then $v = \alpha u$. The geometric meaning of $v = \alpha u$ is that $v$ is a scalar multiple of $u$. In the new problem, with $|x - z| = |x - y| + |y-z|$, as you observed, it's of the same form $|u + v| = |u| + |v|$ with $u = x - y$ and $v = y-z$ (the first thing you tried), so $v = y - z$ is a scalar multiple of $u = x - y$. Think about that for a moment: this means that the line from $y$ to $z$ has the same direction as the line from $x$ to $y$, which already means that $y$ is on the line joining $x$ to $z$: this is already a proof, you are actually done (modulo proving that $y$ is inside the line segment, i.e., $t \in [0, 1]$). But to find an algebraic expression for $t$, you'd go about it as follows.
Note that your result with $u$ and $v$ gives $v = \alpha u$ or here, $y - z = \alpha(x - y)$. Just solve for $y$ in terms of $x$ and $z$! You have
$$ y - z = \alpha(x - y) = \alpha x - \alpha y,$$
so collecting all the $y$ terms on the left-hand side,
$$ \alpha y + y = \alpha x + z$$
$$ (\alpha + 1)y = \alpha x + z$$
and dividing throughout by $(\alpha + 1)$ (because you want the form "$y = $ something"), we get
$$ y = \frac{\alpha}{\alpha + 1}x + \frac{1}{\alpha + 1}z$$
which is of the form you want, with $t = \frac{1}{\alpha + 1}$. No guesswork required; everything worked out fine, the way we knew it would.
When doing these things, there are often (at least) two perspectives: the algebraic and the geometric. It will help you to have in mind the "glossary" between the two, and frequently (at every step if possible) translate back and forth between the two perspectives.
A: The sleek functional analysis solution is $\mathbb{R}^n$ is a Hilbert space $\implies$ $\mathbb{R}^n$ is uniformly convex $\implies$ $\mathbb{R}^n$ is strictly convex $\iff$ your property holds.
