Maximum-depth bound for a brachistochrone ($T\ge [l/\sqrt{h}+\pi \sqrt{h}]/\sqrt{2g}\ge \sqrt{2\pi l/g}$)

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.