For a closed plane curve, showing some inequalities. I have a problem following :
Let $\gamma:[0,T]→\mathbb{R}^2$ be a closed plane curve, i.e., a regular parametrized curve such that $ \gamma$ and all its derivatives agree at 0 and $T$. For convenience of formulation, we assume that $\gamma$ is parametrized by arc-length
(1) Show that there exist $t_1, t_2 \in [0,T]$ such that $\gamma'(t_1)=-\gamma'(t_2)$.
(2) Let $\kappa(t)$ denote the curvature of $\gamma$ at t. Show that $$\int_0^{T} |\kappa(t)|dt \ge 2\pi$$
I tried to solve this problems. Since $\gamma$ is parametrized by arc-length, it is unit-speed curve.  So, I know the part(1) geometrically, but I don't know the perfect answer.
My tutor gave me a hint 'consider the integral $$\int_0^{T} \gamma'(t)dt$$
I know the value of the integral is zero because $\gamma$ is a closed curve. But I don't proceed more..
Also, for part (2), I hope to apply Gauss-Bonnet Theorem. 
Since it is a plane curve, the Gaussian curvature is 0, and the  curvature $\kappa$ is equal to the geodesic curvature $\kappa_g$. But I don't know why does the inequality in (2) hold.
please inform me..
 A: For convenience, let us extend the domain of $\gamma$ to $\Bbb R$ by defining $\gamma(t+nT)=\gamma(t)$, $\forall n\in\Bbb Z$.
(1) Denote the unit circle in $\Bbb R^2$  by $S^1$, i.e. 
$$S^1=\{(\cos \theta, \sin \theta)\in \Bbb R^2\mid \theta\in \Bbb R\}.$$
Since $\gamma$ is  parameterized  by arc-length, $\gamma'$ can be considered as a continuous and $T$-periodic map from $\Bbb R$ to $S^1$. Therefore, $\gamma'(\Bbb R)=\gamma'([0,T])$ must be a closed and connected subset of $S^1$, i.e. there exist $a\le b$, such that 
$$\gamma'(\Bbb R)=\{(\cos \theta, \sin \theta)\in \Bbb R^2\mid \theta\in [a,b]\}.$$
To prove (1), it suffices to show that $b-a\ge\pi$. Denote $\theta=\frac{a+b}{2}$ and $v=(\cos\theta, \sin\theta)$.
$$\int_0^T\langle v,\gamma'(t)\rangle dt= \langle v,\int_0^T\gamma'(t) dt\rangle =0\Longrightarrow \exists t\in[0,T],\, \langle v,\gamma'(t)\rangle\le 0\Longrightarrow  b-a\ge \pi,$$
which completes the proof.

(2) To apply Gauss–Bonnet theorem, let us follow the statements and notations in this page, and apply Equation $(3)$ there. 
If $\gamma$ is a simple closed curve, we may choose $M$ as the closure of the bounded component of $\Bbb R^2\setminus \gamma([0,T])$. Then $\partial M=\gamma([0,T])$, $\chi(M)=1$, $K=0$, $|\kappa_g|=|\kappa|$, and there is no jump angle, so by Gauss–Bonnet theorem,
$$\int_0^T|\kappa(t)|dt\ge |\int_0^T\kappa(t)dt|=2\pi.$$
Otherwise, $\gamma$ is self-intersecting. Let 
$$t_1:=\sup \{t\in[0,T]: \gamma|_{[0,t]} \text{ is a simple curve} \}.$$
Then $0<t_1<T$ and there exists $s_1\in [0,t_1)$, such that $\gamma|_{[s_1,t_1]}$ is a simple closed curve, which has only one posible jump angle $\alpha_1$ at $\gamma(s_1)=\gamma(t_1)$ with $|\alpha_1|\le \pi$. Similar to the simple closed curve case, we may choose $M$ as the closure of the bounded component of $\Bbb R^2\setminus \gamma([s_1,t_1])$ with  $\partial M=\gamma([s_1,t_1])$. By Gauss–Bonnet theorem,
$$\int_{s_1}^{t_1}|\kappa(t)|dt\ge |\int_{s_1}^{t_1}\kappa(t)dt|=|2\pi-\alpha_1|\ge \pi.$$
Similarly, let 
$$t_2:=\sup \{t\in[t_1,s_1+T]: \gamma|_{[t_1,t]} \text{ is a simple curve} \}.$$
Then $t_2>t_1$ and there exists $s_2\in [t_1,t_2)$, such that $\gamma|_{[s_2,t_2]}$ is a simple closed curve, which has only one possible jump angle $\alpha_2$ at $\gamma(s_2)=\gamma(t_2)$ with $|\alpha_2|\le \pi$. Then we may choose $M$ as the closure of the bounded component of $\Bbb R^2\setminus \gamma([s_2,t_2])$ with $\partial M=\gamma([s_2,t_2])$. By Gauss–Bonnet theorem,
$$\int_{s_2}^{t_2}|\kappa(t)|dt\ge |\int_{s_2}^{t_2}\kappa(t)dt|=|2\pi-\alpha_2|\ge \pi.$$
Since $s_1<t_1\le s_2<t_2\le s_1+T$,
$$\int_0^T|\kappa(t)|dt=\int_{s_1}^{s_1+T}|\kappa(t)|dt\ge\int_{s_1}^{t_1}|\kappa(t)|dt+\int_{s_2}^{t_2}|\kappa(t)|dt\ge 2\pi.$$
