= Maximum-depth bound for a brachistochrone
{title2=$T\geq[l/\sqrt h+\pi\sqrt h]/\sqrt{2g}\geq\sqrt{2\pi l/g}$}
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, gives
$$
\frac{ds}{\sqrt y}\geq\frac{dx}{\sqrt h}+\sqrt{\frac{h-y}{hy}}\,|dy|.
$$
The 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> $x=h(\theta-\frac12\sin2\theta)$, $0\leq\theta\leq\pi$. 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.
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