Show $ |k(x_o)| \geq \frac{1}{\|c(x_o)\|} $ , $c$ plane curve and $k(x)$ curvature. Let $ c:(a,b)→\mathbb{R^2}$ be a regular arc length parametrized plane curve. 
Assume that there exists $x_o$ , $a<x_0<b$, such that the function  $x \rightarrow \|c(x)\|$ has a maximum in $x_0$. 
Prove: $$ |k(x_o)| \geq \frac{1}{\|c(x_o)\|} $$ ($k(x)$ curvature)
My Work:
We define a function $f : (a,b) \rightarrow \mathbb{R^2}$ , $ x \rightarrow  \|c(x)\|^2 $.
We get:
$$f'(x) = 2\langle c'(x),c(x)\rangle$$
and
$$ f''(x) = 2( \langle c''(x),c(x)\rangle + \langle c'(x),c'(x)\rangle) = 2(\langle c''(x),c(x)\rangle + 1) = 2(\langle k(x)n(x),c(x)\rangle +1) = 2(k(x)\langle n(x),c(x)\rangle +1) $$
Is this right? ( $k(x)$ curvature , $n(x)$ normal )
f has a maximum in $x_o \Rightarrow f'(x_0) = 0$ and $f''(x_0) \leq 0$. Unfortunately I'm stuck here. Can somebody help me to finish this proof? Maybe we could consider $f''(x_o) = 2(k(x_o)\langle n(x_o),c(x_o)\rangle +1) \leq 0?$
 A: If
$f(x) = \Vert c(x) \Vert^2 = \langle c(x), c(x) \rangle \tag 1$
has a maximum at $x_0$, then
$f'(x) = 2\langle c'(x), c(x) \rangle \tag 2$
must obey
$f'(x_0) = 0, \tag 3$
which by (2) yields
$\langle c'(x_0), c(x_0) \rangle = 0; \tag 4$
i.e.,
$T(x) = c'(x), \tag 5$
the unit tangent vector to $c(x)$, is normal to $c(x)$ at $x_0$:
$\langle T(x_0), c(x_0) \rangle = 0. \tag 6$
Since we are working in the plane $\Bbb R^2$, this forces the unit normal vector $N(x)$ to be collinear with $c(x)$ at $x_0$, for both $c(x_0)$ and $N(x_0)$ are normal (perpendicular) to $T(x_0)$; therefore the expression for $f''(x)$, 
$f''(x) = 2(\langle c''(x), c(x) \rangle + \langle c'(x), c'(x) \rangle) = 2(\langle T'(x), c(x) \rangle + \langle T(x), T(x) \rangle)$
$= 2(\langle \kappa(x)N(x), c(x) \rangle + 1) = 2(\kappa(x)\langle N(x), c(x) \rangle + 1), \tag 7$
may be further reduced at $x_0$ by using the fact that
$c(x_0) = \pm \Vert c(x_0) \Vert N(x_0) \tag 8$
(which is validated in the end note to this answer; see below), with precisely one of the two alternatives actually binding, to obtain
$f''(x_0) = 2(\pm \kappa(x_0)\langle \Vert c(x_0) \Vert N(x_0), N(x_0) \rangle + 1); \tag 9$
since $x_0$ is a local maximum of $\Vert c(x) \Vert^2 = \langle c(x), c(x) \rangle$, we must also have
$f''(x_0) \le 0; \tag{10}$
thus, via (9),
$2(\pm \kappa(x_0)\langle \Vert c(x_0) \Vert N(x_0), N(x_0) \rangle + 1) \le 0, \tag{11}$
whence
$\pm \kappa(x_0)\langle \Vert c(x_0) \Vert N(x_0), N(x_0) \rangle + 1 \le 0, \tag{12}$
whence
$\pm \kappa(x_0)\langle \Vert c(x_0) \Vert N(x_0), N(x_0) \rangle   \le -1; \tag{13}$
at this point we see that in fact the negative sign must in fact be selected, since $\kappa(x) > 0$ and
$\langle \Vert c(x_0) \Vert N(x_0), N(x_0) \rangle = \Vert c(x_0) \Vert \langle N(x_0), N(x_0) \rangle = \Vert c(x_0) \Vert > 0; \tag{14}$
(13) thus becomes
$-\kappa(x_0) \Vert c(x_0) \Vert \le -1, \tag{15}$
or
$\kappa(x_0) \Vert c(x_0) \Vert \ge 1, \tag{16}$
so
$\kappa(x_0) \ge \dfrac{1}{\Vert c(x_0) \Vert}, \tag{17}$
the desired result.
It is of sufficient import to be worth pointing out, I think, that in the course of establishing (17) we have also proven that in fact (see (8))
$ N(x_0) = -\dfrac{c(x_0)}{\Vert c(x_0 \Vert}, \tag{18}$
which shows that $N(x_0)$ is directed oppositely to $c(x_0)$.  Thus the curve $c(x)$ is bending back towards the origin at a local maximum of $\Vert c(x) \Vert$, which is exactly the behavior we should intuitively expect.
Note Added in Edit, Saturday 21 April 2018 2:23 PM PST: This note added, by an large, to address the concerns of our OP RukiaKuchiki which were mentioned in a comment to this answer.  First of all, on the derivation of (8):  once we see that $c(x_0)$ and $N(x_0)$ are co-linear, we know we may write
$c(x_0) = \alpha N(x_0) \tag{18}$
for some $\alpha \in \Bbb R$; then, since $\langle N(x), N(x) \rangle = \Vert N(x) \Vert^2 = 1$, 
$\Vert c(x_0) \Vert^2 = \langle c(x_0), c(x_0) \rangle = \langle \alpha N(x_0), \alpha N(x_0) \rangle = \alpha^2 \langle N(x_0), N(x_0) \rangle = \alpha^2; \tag{19}$
it then follows that
$\alpha = \pm \Vert c(x_0) \Vert; \tag{20}$
substituting this in (18) yields (8).  Second, as far as the sign of $\kappa(x)$ is concerned, we recall that the Frenet-Serret definition of the normal vector $N(x)$ and curvature is ensconced in the equation  
$T'(x) = \kappa(x) N(x); \tag{21}$
by stipulating that $\kappa(x) > 0$ we obtain the convenient formula
$\kappa(x) = \Vert T'(x) \Vert, \tag{22}$
and then $N(x)$ is conveniently the unit vector in the direction of $T'(x)$; so, with these conventions, to which I have here adhered, $\kappa(x) > 0$ by assumption.  End of Note.
