= Solution
For a variation $y+\varepsilon\eta$ with $\eta(a)=0$ but arbitrary $\eta(b)$, differentiation and integration by parts give
$$
\delta I=\int_a^b(F_y\eta+F_{y'}\eta')\,dx
=\left[F_{y'}\eta\right]_a^b+\int_a^b\left(F_y-\frac d{dx}F_{y'}\right)\eta\,dx.
$$
First choose variations supported inside the interval. The fundamental lemma of the <calculus of variations> gives the <Euler-Lagrange equation>. Then allow arbitrary endpoint variation to obtain the <natural boundary conditions for a free endpoint>:
$$
\boxed{\frac d{dx}F_{y'}-F_y=0,\qquad F_{y'}(b)=0.}
$$
Because $F$ has no explicit dependence on $x$,
$$
\frac d{dx}(F-y'F_{y'})=y'F_y+y''F_{y'}-y''F_{y'}-y'\frac d{dx}F_{y'}=0.
$$
Thus the <Beltrami identity> is \b[$F-y'F_{y'}=k$], constant along an extremal.
For the <brachistochrone problem>, put $u=-y>0$, so <conservation of energy> gives the bead speed $v=\sqrt{2gu}$. Its travel time is
$$
T=\frac1{\sqrt{2g}}\int_0^{x_0}\sqrt{\frac{1+u'^2}{u}}\,dx.
$$
The <Beltrami identity> now gives $u(1+u'^2)=2R$ for a positive constant $R$. A descending <cycloid> parametrizes this relation:
$$
x=R(\theta-\sin\theta),\qquad u=R(1-\cos\theta),\qquad u'=\cot(\theta/2).
$$
The free endpoint has $F_{u'}=0$, hence $u'(x_0)=0$. On the first descending arc this occurs at $\theta=\pi$. Therefore
$$
\boxed{R=\frac{x_0}{\pi},\qquad y(x_0)=-\frac{2x_0}{\pi},\qquad
x=\frac{x_0}{\pi}(\theta-\sin\theta),\quad y=-\frac{x_0}{\pi}(1-\cos\theta),\quad0\leq\theta\leq\pi.}
$$
Along this arc $ds=2R\sin(\theta/2)d\theta$ and $v=2\sqrt{gR}\sin(\theta/2)$, so
$$
\boxed{T_{\min}=\pi\sqrt{R/g}=\sqrt{\pi x_0/g}.}
$$
The initial vertical tangent causes no divergent travel time: the quotient $ds/v$ has the finite constant limit $\sqrt{R/g}\,d\theta$.
To verify global minimality, take any admissible wire graph of finite travel time below the release level, and let $H>0$ be its maximum depth. For $0<u\leq H$, the <Cauchy-Schwarz inequality> gives
$$
\frac{\sqrt{1+u'^2}}{\sqrt u}\geq\frac1{\sqrt H}+|u'|\sqrt{\frac1u-\frac1H}.
$$
Indeed the two coefficients on the right, after multiplication by $\sqrt u$, form a unit vector. Since the path starts at depth zero and reaches depth $H$, the weighted total variation is at least the integral over all depths from $0$ to $H$. Consequently
$$
T\geq\frac1{\sqrt{2g}}\left[\frac{x_0}{\sqrt H}+\int_0^H\sqrt{\frac1u-\frac1H}\,du\right]
=\frac1{\sqrt{2g}}\left[\frac{x_0}{\sqrt H}+\frac\pi2\sqrt H\right]
\geq\sqrt{\frac{\pi x_0}{g}}.
$$
The last inequality is minimized at $H=2x_0/\pi$. Our <cycloid> attains equality throughout: its depth increases to $H$ and $u'=\sqrt{H/u-1}$. Thus it achieves the genuine minimum over the wire graphs, not merely a stationary time. Reflection handles a negative horizontal endpoint coordinate by replacing $x_0$ with $|x_0|$.
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