Any segment can't be inside the sphere I'm trying to prove if two points $a$ and $b$ are in the closed ball, then the segment between them is inside the ball, and can't be in the sphere, in another words:
Let $a,b\in \mathbb R^n$, $a\neq b$ with the norm induced by the inner product. If $|a|\le r$ and $|b|\le r$, then $|(1-t)a+tb|\lt r$, for every $t\in (0,1)$.
My attempt of solution:
$|(1-t)a+tb|\le (1-t)|a|+t|b|\le (1-t)r+tr= r$ 
But why does this inequality is restrict ($\lt$)?
Thanks in advance
 A: If $a$ is not in the sphere, then $\|a\| < r$ so the desired inequality follows from what you've written; likewise if $b$ is not in the sphere. Therefore, suppose both $a$ and $b$ are in the sphere and are distinct; if $a = b$, the line between them is degenerate.
If $\|(1-t)a+tb\| < (1-t)\|a\| + t\|b\|$, then you're done, so you just need to deal with the case where $\|(1-t)a+tb\| = (1-t)\|a\| + t\|b\|$; i.e. when the triangle inequality gives an equality. Luckily, we know exactly when this happens: $\|x + y\| \leq \|x\| + \|y\|$ with equality if and only if one of $x$ or $y$ is a non-negative scalar multiple of the other. Therefore, $(1 - t)a = \lambda tb$ and for any $t \neq 0, 1$, we see that $a = kb$ where $k = \frac{\lambda t}{1-t}$. As $a$ and $b$ lie on the sphere, the only possible values of $k$ are $\pm 1$ (you should check this), but as $a \neq b$, we see that $a = -b$. But then 
$$\|(1-t)a + tb\| = \|(t-1)b+tb\| = \|(2t - 1)b\| = |2t-1|\|b\| = |2t - 1|r.$$
As $-1 < 2t - 1 < 1$ for $t \in (0, 1)$, we're done.
Added Later: Above I claimed that $\|x+y\| = \|x\| + \|y\|$ if and only if one of $x$ or $y$ is a non-negative scalar multiple of the other. This is not true in general, but is true for a norm of a normed vector space over $\mathbb{R}$ induced by an inner product. More generally, it is true if and only if the normed space is strictly convex, i.e. the unit ball is strictly convex. What this exercise shows is that if the norm of a normed space is induced by an inner product, then the normed space is a strictly convex space.
Note, this may seem like circular reasoning upon first glance, but it isn't. If the norm on a normed vector space over $\mathbb{R}$ is induced by an inner product there is a proof of the fact "$\|x+y\| = \|x\|+\|y\|$ if and only if one of $x$ or $y$ is a non-negative scalar multiple of the other" which is independent of the strictly convex hypothesis. The proof goes as follows:

If $x$ is the zero vector, then $\|x + y\| = \|x\| + \|y\|$ and $x = 0y$; likewise if $y = 0$. Suppose then that $x$ and $y$ are not the zero vector. Then $$\|x+y\|^2 = \langle x+ y, x+y\rangle = \langle x,x\rangle + \langle x,y\rangle + \langle y, x\rangle + \langle y, y\rangle = \|x\|^2 + 2\langle x, y\rangle + \|y\|^2.$$ On the other hand, if $\|x + y\| = \|x\| + \|y\|$, then $$\|x+y\|^2 = \|x\|^2+2\|x\|\|y\|+\|y\|^2.$$ Comparing the two expressions for $\|x+y\|^2$, we see that $\langle x, y\rangle = \|x\|\|y\|$. As $\langle x, y\rangle = \|x\|\|y\|\cos\theta$, we see that $\theta = 0$, i.e. $x$ and $y$ lie on the same ray, so $x = \lambda y$ for some $\lambda > 0$. Conversely, if $x = \lambda y$ for $\lambda > 0$, then $$\|x + y\| = \|x + \lambda x\| = |1+\lambda|\|x\| = \|x\| + \lambda\|x\| = \|x\| + \|\lambda x\| = \|x\| + \|y\|.$$

