Hamilton's equations from variational principle

As we used to derived the Lagrange's equations of motion from variational principle i.e.Hamilton's principle, we can derive Hamilton's equations. 

Hamilton's principle i.e. variational principle gives the condition that, $$ \delta \int_{t_1}^{t_2} L \, dt = 0 $$ 
where L is the Lagrangian of the given system. In the previous derivation, it was considered various paths in the configuration space. Where else, now we will working in the paths which is the part of Phase space where q and p are the independent coordinates.
q - position and 
p - momentum

To go from Lagrangian to Hamilton, we just express the Lagrangian in terms of Hamiltonian where everything is the function of q and p.

$$ \delta \int_{t_1}^{t_2} \left(p_i\dot{q_i} - H (q,p,t)\right) dt = 0 $$  
Let us call this new function as L',
therefore, $$ L' = p_i\dot{q_i} - H $$ and this new L' obeys the Lagrangian equations of motion that is derived earlier from the definition. 
Therefore, $$ \delta \int_{t_1}^{t_2} L'(q, \dot{q}, p, \dot{p}) \,dt = 0 $$
where the integral is defined over 2n dimensional phase space. And it gives symmetrically two sets of relations,
$$ \frac{d}{dt}\left(\frac{\partial{L'}}{\partial\dot{q_j}}\right) - \frac{\partial{L'}}{\partial{q_j}} = 0  \,\,\,\,j = 1,2,...n $$ and
$$ \frac{d}{dt}\left(\frac{\partial{L'}}{\partial\dot{p_j}}\right) - \frac{\partial{L'}}{\partial{p_j}} = 0 \,\,\,\, j = 1,2,....n $$ 
Substituting for L' we get, $$ \frac{\partial{H}}{\partial{q_j}} = -\dot{p_j} $$  and $$ \frac{\partial{H}}{\partial{p_j}} = \dot{q_j} $$ 
which is our desired Hamilton's equations of motion. 
The derivation is nothing but just the result of our Legendre transformation in variational principle.