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A cycloid is the plane curve traced by a point on a circle rolling without slipping along a straight line. A parametrization with rolling radius $R$ is $x=R(\theta -\sin \theta )$, $y=R(1-\cos \theta )$; reversing the sign of $y$ reflects the curve. Its cusps occur at integer multiples of $2\pi$. A descending cycloidal arc solves the classical brachistochrone problem.

Consider a frictionless track from rest between equal-height endpoints separated horizontally by $l>0$. Let depth $y$ be positive downward and let the maximum depth be $h>0$. Conservation of energy gives speed $\sqrt{2gy}$. The Cauchy-Schwarz inequality, applied to the tangent displacement, givesThe squared norm of the coefficient vector is $1/h+(h-y)/(hy)=1/y$. The track goes from depth zero to $h$ and returns to zero, so the second term integrates to at least $2{\int }_0^h\sqrt{(h-y)/(hy)}dy=\pi \sqrt{h}$. Integrating the first term gives $l/\sqrt{h}$, even if the track backtracks horizontally. Division by $\sqrt{2g}$ proves the first bound. The arithmetic-geometric mean inequality proves the second.

Equality requires $h=l/\pi$ and tangent slope $|dy/dx|=\sqrt{(h-y)/y}$, on the descending and ascending halves respectively. Substitution $y=h{\sin }^2\theta$ integrates to the cycloid $x=h(\theta -\frac{1}{2}\sin 2\theta )$, $0\le \theta \le \pi$. It attains the bound, proving global minimality of the single-arch brachistochrone problem between these endpoints. Endpoint singularities are handled as limits of the integrals; the cycloid has finite travel time.

For a variation $y+\epsilon \eta$ with $\eta (a)=0$ but arbitrary $\eta (b)$, differentiation and integration by parts giveFirst 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:Because $F$ has no explicit dependence on $x$,Thus the Beltrami identity is $F-y^{\prime }{F}_{y^{\prime }}=k$, 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 isThe Beltrami identity now gives $u(1+u^{\prime 2})=2R$ for a positive constant $R$. A descending cycloid parametrizes this relation:The free endpoint has $F_{u^{\prime }}=0$, hence $u^{\prime }(x_0)=0$. On the first descending arc this occurs at $\theta =\pi$. ThereforeAlong this arc $ds=2R\sin (\theta /2)d\theta$ and $v=2\sqrt{gR}\sin (\theta /2)$, soThe 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 $0<u\le H$, the Cauchy-Schwarz inequality givesIndeed 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 $H$. ConsequentlyThe last inequality is minimized at $H=2x_0/\pi$. Our cycloid attains equality throughout: its depth increases to $H$ and $u^{\prime }=\sqrt{H/u-1}$. 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 $|x_0|$.

For fixed endpoint values, stationarity requires the Euler-Lagrange equation $f_y-d(f_{y^{\prime }})/dx=0$. Differentiate $y^{\prime }{f}_{y^{\prime }}-f$ along an extremal: its derivative is $y^{\prime }[d(f_{y^{\prime }})/dx-f_y]-f_x$. Thus when $f_x=0$ the Beltrami identity makes that quantity constant.

Take $y$ as depth measured downward. Conservation of energy from rest gives $m{v}^2/2-mgy=0$, hence speed $v=\sqrt{2gy}$. With arc length $ds=\sqrt{1+y^{\prime 2}}dx$, the travel time is $T=(2g{)}^{-1/2}{\int }_0^{x_B}\sqrt{(1+y^{\prime 2})/y}dx$. For $f=\sqrt{(1+y^{\prime 2})/y}$, the Beltrami quantity is $-1/\sqrt{y(1+y^{\prime 2})}$. Thus $y(1+y^{\prime 2})=c>0$.

Put $y=c{\sin }^2\theta$ and take $y^{\prime }=\cot \theta$ along the first arch. Since $dy/d\theta =2c\sin \theta \cos \theta$, one obtains $dx/d\theta =2c{\sin }^2\theta$. Integrating from the initial cusp gives the cycloid solving the brachistochrone problem:For the equal-height target, the first complete arch has ${\theta }_B=\pi$, so $c=l/\pi$. Also $ds/d\theta =2c\sin \theta$ and $v=\sqrt{2gc}\sin \theta$ for $0<\theta <\pi$. ThereforeThe endpoint zeros of speed cancel the zeros of $ds/d\theta$, leaving finite travel time. Global minimality follows from the maximum-depth bound for a brachistochrone: any competing track with maximum depth $h$ satisfies $T\ge [l/\sqrt{h}+\pi \sqrt{h}]/\sqrt{2g}\ge \sqrt{2\pi l/g}$. The displayed cycloid attains both inequalities with $h=l/\pi$.