= Solution
Assume the integrand is sufficiently <smooth> near the positive admissible profiles. Define the <second variation> as $\delta^2F[y,\xi]=\left.d^2F[y+\varepsilon\xi]/d\varepsilon^2\right|_{\varepsilon=0}$, so it is twice the quadratic Taylor coefficient. Differentiation under the integral gives
$$
\delta^2F=\int_\alpha^\beta\left(f_{yy}\xi^2+2f_{yp}\xi\xi'+f_{pp}\xi'^2\right)dx,\qquad p=y'.
$$
Integrating the middle term by parts, with $\xi(\alpha)=\xi(\beta)=0$, yields
$$
\boxed{\delta^2F=\int_\alpha^\beta\left[\left(f_{yy}-\frac d{dx}f_{yp}\right)\xi^2+f_{pp}\xi'^2\right]dx}.
$$
All coefficients are evaluated along $y$.
For the surface-area integrand $f=2\pi y\sqrt{1+p^2}$, the <Beltrami identity> for a stationary profile reads
$$
f-pf_p=\frac{2\pi y}{\sqrt{1+y'^2}}=2\pi E.
$$
Equivalently, its <Euler-Lagrange equation> is $yy''=1+y'^2$, solved by the positive <catenoid> profile $y=E\cosh((x-x_0)/E)$ with $E>0$. Equal boundary radii at $x=\pm L$, for $L>0$, require $x_0=0$, so
$$
\boxed{y(x)=E\cosh(x/E),\qquad a=E\cosh(L/E)}.
$$
Direct substitution also verifies the <Euler-Lagrange equation>. This is a stationary profile whenever a positive $E$ satisfying the boundary relation exists; such an $E$ does not exist for arbitrary ring separations and radii.
On this profile, with $z=x/E$, the coefficients are $f_{yy}=0$, $f_{pp}=2\pi E\operatorname{sech}^2z$ and $d f_{yp}/dx=(2\pi/E)\operatorname{sech}^2z$. Changing variables therefore gives
$$
\boxed{\delta^2F=2\pi\int_{-L/E}^{L/E}(\xi_z^2-\xi^2)\operatorname{sech}^2z\,dz}.
$$
The printed strict inequality must mean \b[every nonzero admissible variation]: the zero function makes the integral zero, so the literal quantifier including it is false.
For completeness, the strict positivity here is sufficient for a strict local minimum in the fixed-endpoint $C^1$ topology, among positive axisymmetric graph profiles. On this finite interval the weight $p(z)=\operatorname{sech}^2z$ has a positive minimum. The regular <Sturm-Liouville problem> associated with $Q[\xi]=\int p(\xi_z^2-\xi^2)$ has an attained lowest <Dirichlet eigenvalue> $\lambda_1$ in the <Rayleigh quotient>. Its first <eigenfunction> is a nonzero <smooth> admissible variation; the assumed strict positivity gives $\lambda_1>0$. Hence $Q\geq\lambda_1\|\xi\|_{L^2}^2$. Combining this with $Q\geq p_{\min}\|\xi_z\|_{L^2}^2-p_{\max}\|\xi\|_{L^2}^2$ proves <coercivity> in the <Sobolev space> $H_0^1$.
The coefficients of the <second variation> depend continuously on $(y,y')$ while $y>0$. For a sufficiently small $C^1$ perturbation, the quadratic forms along the segment from the stationary profile to the perturbed one retain a uniform positive <coercivity> bound. Taylor's formula with integral remainder, and the vanishing <first variation>, then gives $F[y+v]-F[y]\geq c\|v\|_{H^1}^2>0$ for nonzero sufficiently small admissible $v$. \b[The stationary area is therefore a strict local minimum under the stated nonzero-variation condition]. This supplies the uniform bound needed in an infinite-dimensional problem, rather than relying on pointwise positivity alone in an arbitrary function space.
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