Geodesics of a Sphere in Cartesian Coordinates I want to minimize $I = \int |\dot{x}|^2 dt$
subject to the constraint $|x|^2=1$ (sphere)
which gives an Euler equation of $\lambda x - \ddot{x} = 0$.
I have to show that the Euler equation is actually $|\dot{x}|^2 x - \ddot{x} = 0$.
Is it right to assume that $\lambda=|\dot{x}|^2$ simply by the fact that it minimizes $I^* = \int |\dot{x}|^2- \lambda (|x|^2-1) dt$ which is $\geq 0$, so the $\lambda$ that minimizes $I^*$ is $|\dot{x}|^2$?
If I then try to integrate the Euler equation, then I get a SHM equation:
$x1= A \cos(|\dot{x}| t - C)$ where A, C are constants
and similarly for x2, x3
But how do I combine them to give the equation of a great circle, since I don't know the $C$'s?
Thank you for any enlightenment! 
 A: Firstly, your expression for $I^*$ does not agree with your Euler equation: if you write $I^*$ as you did, the Euler equation should be $\ddot{x} + \lambda x = 0$. (In fact, you have a general sign error. The correct Euler equation for the geodesic is $\ddot{x} + |\dot{x}|^2 x = 0$, if the sign were as you wrote, the solution would not be a trigonometric function, but rather an exponential function.)
Now, to compute $\lambda$, you need to use the constraint $|x|^2 = 1$ twice. First, take the dot product of the Euler equation with $x$, you get that
$$ x\cdot \ddot{x} + \lambda |x|^2 = x \cdot\ddot{x} + \lambda = 0$$
Second, take the second time derivative of the constraint
$$ 0 = \frac{d^2}{dt^2}(|x|^2 - 1) = \frac{d}{dt} (x\cdot \dot{x}) = \dot{x}\cdot\dot{x} + x \cdot \ddot{x} $$
Comparing the two equations and solving for $\lambda$, you have that $\lambda = |\dot{x}|^2$. Hence the correct Euler equation is in fact
$$ \ddot{x} + |\dot{x}|^2 x = 0 $$

For your second question, the $A$s and $C$s (a total of 6 free variables) are fixed by the initial data: the initial position and initial velocity of the geodesic. In other words, you have that 
$$ x_i(t) = A_i \cos\left( |\dot{x}|^2 t - C_i\right) $$
and you want to solve for $A_i, C_i$ for prescribed values $x_i(0)$ and $\dot{x}_i(0)$. (Of course, the initial data must satisfy the constraints that the velocity vector is orthogonal to the position vector, and that the position vector has norm 1.)
A: Let the initial data be given by:
$x(t=0) = x_0,\quad\dot{ x}(t=0) = v_0  $
The equations of motion of the constrained Lagrangian define a simple harmonic motion. After the substitution of the initial data, the solution has the form
$x = x_0 \cos(\sqrt{\lambda} t) + \frac{ v_0}{\sqrt{\lambda}} \sin(\sqrt{\lambda} t) $
The requirement that at all times $t$
$|x|^2 = 1$
Implies (by the substitution of the solution into the constraint equation)
$|x_0|^2 = 1,  v_0. x_0 = 0, | v_0|^2 = \lambda$
Therefore (again by substitution)
$|\dot{x}(t)|^2 = \lambda = |v_0|^2$
Therefore 
$\lambda = |\dot{ x}(t)|^2$
To see that the trajectory is a great circle, we notice that it lies on the plane (passing through the sphere's center): 
$ w. x(t)= 0$
Where: 
$\vec w = x_0 \times  v_0$
