TITLE: Intuition behind variational principle
QUESTION [6 upvotes]: Hofer-Zehnder, in section 1.5, proves that every Hamiltonian field on a strictly convex compact regular energy surface carries a periodic orbit. I have understood the proof. What I am wondering about is how he gets the variational principles. The start of the proof finds a function $H$ having the given convex hypersurface as energy surface and satisfying positive homogeneity of degree two, positive definiteness of the Hessian, and $\mathcal{C}^2$ regularity. Then the proof normalizes the period to get period $2\pi$, and then it goes:

For example, the variational principle:
$$\min\int_0^{2\pi}H\big(x(t)\big)\,dt\quad{\rm under}\quad\frac{1}{2}\int_0^{2\pi}\langle Jx,\dot x\rangle=1.$$
[…] has the above equations as Euler equations.

Firstly, what are the Euler equations of a variational principle? And secondly, how did they get this particular principle?
Update
Let me make my question more precise. There are 3-4 points I have trouble with.

Why do they consider this principle with this strange constraint of that integral being 2? I mean, it isn't even a vector space! It's perhaps not even an affine space, it has no particular properties. Why should this constraint be useful? And why that functional?
How do I get the Euler equations for this principle, and what are they? If I try to say "directional derivatives 0", I get stuck at once, because how am I supposed to make the variation?
The book then goes on, and says this functional attains neither the infimum, nor the supremum. How can one prove these claims?
The book then Legendre transforms $H$ into $G$, gives properties of $G$, and introduces the same principle on $G$:

$$\min\int_0^{2\pi}G(\dot z)\,dt\quad{\rm under}\quad\frac{1}{2}\int_0^{2\pi}\langle Jz,\dot z\rangle=1.$$

Again, why should I consider this particular principle, with this particular constraint? I mean, I understood the rest of the proof so I know this is the right principle to consider, but how was it come up with? Is there any reason that would make one think of this principle as a possible correct choice?

REPLY [3 votes]: Let there be given a curve $\vec{p}(t)$ and a real valued function $L$ with the following arguments: this curve, the time derivative of the curve and the time $t$ itself. Now consider the following integral:
$$
W\left(\vec{p},\dot{\vec{p}}\right) = \int_{t_1}^{t_2} L\left(\vec{p},\dot{\vec{p}},t\right) dt
$$
The integral $W$ is dependent upon the chosen curve $\vec{p}(t)$ ; it is a so-called functional of $\vec{p}$ and $\dot{\vec{p}}$.
We wish to determine for which curves $\vec{p}(t)$ the functional $W(\vec{p},\dot{\vec{p}})$ has an extreme value. But it is not possible to simply take a derivative and put it to zero. So we must do something else. Assume that the extreme value is obtained for the curve $\vec{q}(t)$ and let
$\vec{p}(t) = \vec{q}(t) + \epsilon \vec{f}(t)$ , with the same end-points 
$\vec{p}(t_1) = \vec{q}(t_1)$ and $\vec{p}(t_2) = \vec{q}(t_2)$, hence
$\vec{f}(t_1) = \vec{f}(t_2) = \vec{0}$. For the rest, the curve $\vec{f}(t)$ is arbitrary and the variable $\epsilon$ does not depend on $t$. In this way our functional $W$ has become a common function of $\epsilon$:
$$
W\left(\vec{p},\dot{\vec{p}}\right) =
\int_{t_1}^{t_2} L\left(\vec{q} + \epsilon \vec{f},\dot{\vec{q}} + \epsilon \dot{\vec{f}},t\right) dt
$$
A common real-valued function, which can be differentiated in the common way to find extremes:
$$
\frac{dW}{d\epsilon} = 0 \qquad \Longleftrightarrow \qquad
\int_{t_1}^{t_2} \frac{dL\left(\vec{q} + \epsilon \vec{f},\dot{\vec{q}} + \epsilon \dot{\vec{f}},t\right)}
{d\epsilon} dt \quad \mbox{for} \; \epsilon = 0
$$
Let the (vector) components of $\,\vec{p}\,$ be $\,p_k$ . Apply the chain rule:
$$
\frac{dL}{d\epsilon} = \sum_k \frac{\partial L}{\partial p_k} \frac{d p_k}{d\epsilon}
+ \sum_k \frac{\partial L}{\partial \dot{p}_k} \frac{d \dot{p}_k}{d\epsilon}
+ \frac{\partial L}{\partial t} \frac{d t}{d\epsilon} = 
\sum_k \frac{\partial L}{\partial p_k} f_k + \sum_k \frac{\partial L}{\partial \dot{p}_k} \dot{f}_k + 0
$$
The last term can be rewritten with help of:
$$
\frac{d}{dt} \left\{ \frac{\partial L}{\partial \dot{p}_k} f_k\right\} =
\frac{\partial L}{\partial \dot{p}_k} \dot{f}_k +
\frac{d}{dt} \left(\frac{\partial L}{\partial \dot{p}_k}\right) f_k
$$
It follows that:
$$
\frac{dW}{d\epsilon} = \sum_k \int_{t_1}^{t_2} \left[\frac{\partial L}{\partial p_k} f_k
- \frac{d}{dt} \left(\frac{\partial L}{\partial \dot{p}_k}\right) f_k
+ \frac{d}{dt} \left\{ \frac{\partial L}{\partial \dot{p}_k} f_k\right\} \right] dt = \\
\sum_k \int_{t_1}^{t_2} \left[\frac{\partial L}{\partial p_k}
- \frac{d}{dt} \left(\frac{\partial L}{\partial \dot{p}_k}\right) \right] f_k\, dt +
\sum_k \left[ \frac{\partial L}{\partial \dot{p}_k} f_k \right]_{t_1}^{t_2}
$$
But we know that $f_k(t_1) = f_k(t_2) = 0$ , so the last term is zero. Furthermore we know that, for $\epsilon = 0$ : $p_k = q_k$ , and the functions $f_k$ are completely arbitrary. Therefore
it is concluded that the integral is zero if the folowing necessary conditions (not sufficient
eventually) are fulfilled for all $k$ :
$$
\frac{\partial L}{\partial q_k} - \frac{d}{dt} \left(\frac{\partial L}{\partial \dot{q}_k}\right) = 0
$$
These are the well known Euler-Lagrange equations. It is noted that special notations like $\delta W$ for the "variation" of the integral are not needed. The variation is reduced to common differentiation, which makes the (IMHO confusing) $\delta$ practice completely redundant.