In recent Calculus of Variations lecture, I learnt about the Brachistochrone problem and its solution. For those who don't know, it is a standard problem in dynamics which is often used as a motivating example in introductions to functional analysis. It goes as follows:
$\textbf{The problem}$
Suppose that a bead slides on a frictionless wire in a vertical plane. What shape of the wire minimises the time for the bead to fall from rest at a point $A$ to a lower, and horizontally displaced, point $B$?
$\textbf{Textbook solution}$
Choose $A$ to be the origin of coordinates in the vertical plane with x being horizontal distance from the origin and y being the distance below the origin. The bead starts with zero velocity so conservation of energy implies that its speed $v$ at any later time is given by $v=\sqrt{2yg}$, where $g$ is the acceleration due to gravity. Hence we have to minimise
$$T=\int_{A}^{B}\frac{dl}{v}=\frac{1}{\sqrt{2g}}\int_{A}^{B}\frac{\sqrt{dx^2+dy^2}}{\sqrt{y}}$$
For simplicity, let us assume that $x$ is a good coordinate on the curve, so that $y=y(x)$ and
$$T=\int_{0}^{x_B}\sqrt{\frac{1+(y')^2}{y}}dx$$
To extremize $T$, we can use the Euler-Lagrange equations. As the integrand has no explicit $x$ dependence, these equations reduce to
$$C=\sqrt{\frac{1+(y')^2}{y}}-y'\frac{\partial}{\partial y'}\left(\sqrt{\frac{1+(y')^2}{y}}\right)=\frac{1}{\sqrt{y[1+(y')^2]}}$$
where $C$ is a constant. The resulting differential equation is $y[1+(y')^2]=2C$, where $C$ is a positive constant. This is well-known to describe a cycloid, i.e. the path traced by a point on the rim of a circle as the circle rolls along a straight line without slipping. It can further be show that the solution is given parametrically by $x=C(\theta-\sin\theta)$, $y=C(1-\cos\theta)$.
Now, this is where I am a little puzzled. All we have shown so far is that the functional $T$ is stationary when $y$ is a cycloid, and yet every single source I have found online claims that $y$ is an $\textit{absolute minimum}$ of $T$. Is this due to a misunderstanding on my part, or is this assertion indeed non-trivial and needs proving?
