= Catenary minimizes the integral of height along a fixed-length curve
{title2=$\min\int y\,ds$}
Among <curves> of fixed <arc length> $L>2a$ joining $(-a,0)$ to $(a,0)$, the hanging <catenary> globally minimizes $\int y\,ds$. Parametrize a competitor by <arc length> $s\in[-L/2,L/2]$, so $x'^2+y'^2=1$. <Integration by parts> and the fixed horizontal endpoint separation give, for every $b>0$,
$$
\int_{-L/2}^{L/2}y\,ds=-\int_{-L/2}^{L/2}sy'\,ds
=2ab-\int_{-L/2}^{L/2}(bx'+sy')\,ds
\geq2ab-\int_{-L/2}^{L/2}\sqrt{b^2+s^2}\,ds.
$$
The inequality is the <Cauchy-Schwarz inequality>. Equality requires $x'=b/\sqrt{b^2+s^2}$ and $y'=s/\sqrt{b^2+s^2}$. <Integration> gives the <catenary>
$$
y=b\left[\cosh(x/b)-\cosh(a/b)\right],\qquad L=2b\sinh(a/b).
$$
The last equation has a unique positive $b$, since $\sinh q/q$ increases strictly for $q>0$. The equality <curve> is a single-valued graph below the chord, so the same global bound proves the result under those additional restrictions.
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