A transition from Riemannian geometry to Symplectic geometry I have tried to prove that exercises 1&2 on the page 27, Lectures on Symplectic Geometry, Ana Cannas da Silva, which is available on professor's website: https://people.math.ethz.ch/~acannas/Papers/lsg.pdf
Let $(X, g)$ be a Riemannian manifold. The arc-length of a smooth curve $ \gamma : [a, b] \to X$ is 
$$l(\gamma)= \int_{a}^{b} \left|\frac{d\gamma}{dt}\right| dt,$$
where $\left\vert\frac{d\gamma}{dt}\right\vert= \sqrt{g_{\gamma(t)}\left(\frac{d\gamma}{dt},\frac{d\gamma}{dt}\right)}.$
A curve $\gamma$ is called a curve $\textit{of constant velocity}$ when $\left|\frac{d\gamma}{dt}\right|$ is independent of $t$.

Questions: 1) Show that, given any curve $ \gamma : [a, b] \to X$ (with $\frac{d\gamma}{dt}$ never vanishing), there is a reparametrization $\tau: [a,b] \to [a,b]$ such that $\gamma \circ \tau: [a,b] \to X$ is of constant velocity.
2) Given a smooth curve $ \gamma : [a, b] \to X$, the action $$\mathcal{A}(\gamma):= \int_{a}^{b} \left|\frac{d\gamma}{dt}\right|^2 dt. \tag {1}$$ Show that, among all curves joining x to y, $\gamma$ minimizes the action if and only if $\gamma$ is of constant velocity and $\gamma$ minimizes arc-length.

Hint(2): Suppose that $\gamma$ is of constant velocity, and let $\tau: [a,b] \to [a,b]$ be a reparametrization. Show that $\mathcal{A}(\gamma \circ \tau) \geq \mathcal{A}(\gamma)$, with equality only when $\tau=id$.
I haven't discovered possible patterns of proof yet. Any hints or suggestions?
As a consequence, a curve $ \gamma : [a, b] \to X$ is called a $\textit{geodesic}$ if it locally minimizes arc-length and is of constant velocity. Then by using this geodesic one can construct a self symplectomorphism of tangent bundle that is just a $\textit{geodesic flow}.$
 A: Let $\Gamma : (-\varepsilon, \varepsilon) \times [a,b] \to X$ be a proper variation of $\gamma$, given by $\Gamma(s,t) = \exp (s V(t))$, where $V$ is the proper variation field of $\Gamma$ defined as $V(t) = \partial_s \Gamma(0,t)$, with $V(a) = V(b) = 0$ along $\gamma$. 
Then the action $(1)$ is given by the variation of energy 
$$\mathcal A (\gamma) = \int_a^b g \left(\partial_t \Gamma_s (t), \partial_t \Gamma_s (t)\right) dt \tag{2} $$
Differentiating (2) under the integral sign with respect to $s$ we get 
$$\begin{aligned} \frac{d\mathcal A}{ds}(\gamma)&= \int_a^b \frac{\partial}{\partial s}\bigg[ g_{\gamma(t)}(\partial_t \Gamma_s(t), \partial_t \Gamma_s (t)) \bigg]dt \\&\underset{(1)}= \int_a^b g_{\gamma(t)}(D_s\partial_t \Gamma_s(t), \partial_t \Gamma_s(t)) + g_{\gamma(t)}(\partial_t \Gamma_s(t),D_s \partial_t \Gamma_s(t)) dt \\& \underset{(2)}= 2 \int_a^b g_{\gamma(t)}(D_s \partial_t \Gamma_s(t), \partial_t \Gamma_s(t)) dt \\&\underset{(3)}=2 \int_a^b g_{\gamma(t)}(D_t \partial_s \Gamma_s(t),\partial_t  \Gamma_s(t))\\&\underset{(4)}=2 \int_a^b \frac{d}{dt}\bigg[g_{\gamma(t)}(\partial_s \Gamma_s(t), \partial_t \Gamma_s(t))\bigg]dt - 2\int_a^b g_{\gamma(t)}(\partial_s \Gamma_s(t), D_t\partial_t \Gamma_s(t))dt \\& = 2\ g_{\gamma(t)}( \partial_s \Gamma_s(t), \partial_t \Gamma_s(t))\bigg|_a^b - 2\int_a^b g_{\gamma(t)}(\partial_s \Gamma_s(t), D_t\partial_t \Gamma_s(t))dt\\&= 2\  g_{\gamma(t)}( \partial_s \Gamma_s(b), \partial_t \Gamma_s(b)) -2\  g_{\gamma(t)}( \partial_s \Gamma_s(a), \partial_t \Gamma_s(a)) \\& \quad - 2\int_a^b g_{\gamma(t)}(\partial_s \Gamma_s(t), D_t\partial_t \Gamma_s(t))dt
\end{aligned}$$
where we have used that, (1) $g_{\gamma}$ is symmetric , (2) $D_s\partial_t \Gamma = D_t \partial_s \Gamma$ (Symmetry Lemma), (1) and (4) product rule for inner product. 
At $s = 0$ we obtain 
$$\begin{aligned}\frac{d\mathcal A}{ds}\bigg|_{s=0}(\gamma) &=2\  g_{\gamma(t)}( V(b), \dot{\gamma}(t)) -2\ g_{\gamma(t)}( V(a), \dot\gamma (t)) \\& \quad - 2\int_a^b g_{\gamma(t)}(V(t), D_t\dot\gamma (t))dt  \\& = -2\int_a^b g_{\gamma(t)}(V(t), D_t\dot\gamma (t))dt
\end{aligned}$$
Thefore $\frac{d\mathcal A}{ds}\bigg|_{s=0}(\gamma) = 0$ if and only if $D_t \dot \gamma (t) = 0$, that is, if and only if $\gamma$ is a geodesic. 
A: Actually, there is an easier solution via Holder's inequality:
Denote $f(t) = |\gamma'(t)|$, then
$$l(\gamma)^2 = \left(\int_a^b 1\cdot f(t) dt\right)^2 \leq \left(\int_a^b f^2(t)dt\right)\left(\int_a^b 1^2 dt\right)= (b - a)\mathcal{A}(\gamma)$$
Here I used the Holder's inequality for $p = q = 2$ for $f$ and $1$.
It's known that the equality holds iff $f\equiv const$ and it gives us the solution.
