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Write $s=A(S)=4\pi$, $L=L(\partial U)$, and $a=\min \{A(U),A(S\setminus U)\}$. We first prove a constant-version spherical isoperimetric inequality directly, including regions with several boundary components.

If $L\ge 1$, the arithmetic bound $A(U)A(S\setminus U)\le s^2/4=4{\pi }^2$ already gives $A(U)A(S\setminus U)\le 4{\pi }^2{L}^2$. If $L=0$, a region with smooth boundary is empty or the whole connected sphere up to boundary-null changes, and its area product is zero. For $0<L<1$, write the smooth boundary as finitely many disjoint simple closed curves ${\gamma }_j$ of lengths $l_j$. Fix a point $p_j$ on each curve. The shorter boundary arc from $p_j$ to any other point has length at most $l_j/2$, so ${\gamma }_j$ is contained in the closed geodesic ball of radius $l_j/2$ about $p_j$.

The complement of that small ball is connected and lies entirely on one side of ${\gamma }_j$. The Jordan curve theorem therefore gives a disk $E_j$ bounded by ${\gamma }_j$ contained in the ball. Its area satisfiesThe indicator function of $U$ and the sum ${\sum }_j{1}_{E_j}$ have the same jumps modulo two across every boundary curve. Their difference modulo two is consequently constant throughout the sphere off the boundary. Thus either $U$ or its complement lies in ${\bigcup }_j{E}_j$, up to its area-zero boundary. This also handles nested curves and disconnected regions. It follows thatCombining the cases, one valid universal choice is $\kappa =4{\pi }^2$. Sharpness is not required.

For the inclusion estimate put $x=A(V)$, $d=A(D)>0$ and $y=A(V\cap D)$. Splitting $D$ into its portions inside and outside $V$ givesBecause $0\le A(D\setminus V)\le s-x$ and $0\le A(D\cap V)\le x$, this expression lies between $-x(s-x)$ and $x(s-x)$. Apply the preceding spherical isoperimetric inequality to $V$ and divide by $sd$:Thus the constant in the spherical inclusion area discrepancy is independent of the nonempty open set $D$.

Let $g$ and $\tilde{g}$ be the old and new smooth Riemannian metrics. Their area forms satisfyThe determinant ratio is invariant under changes of coordinates, so $\theta$ is a globally defined smooth function. The layer cake representation follows from $\theta (x)={\int }_0^{\infty }{1}_{\{\theta (x)>t\}}dt$ and Tonelli's theorem:Smoothness makes every $D_t$ open. Compactness of the sphere gives $0<m=\min \theta \le M=\max \theta <\infty$. Multiply the inclusion bound by $A(D_t)$ before integrating; for empty $D_t$ the resulting bound is trivially zero. For $t<m$, $D_t=S$ and the discrepancy is exactly zero; for $t>M$, $D_t$ is empty. ThereforeDividing by $\tilde{A}(S)>0$ provesThis constant depends only on the two metrics. If $\theta$ is constant the normalized areas agree exactly, as the bound also shows.

The preceding parts show that an area maximizer among curves of fixed length is a plane section of the sphere, hence a circle. If its smaller cap has angular radius $0\le \theta \le \pi /2$, thenFor this circle,No curve of the same length encloses more area than the maximizing circle. Equivalently every curve enclosing the smaller area $A$ satisfies the spherical isoperimetric inequalitywith equality precisely for circles.

On the unit sphere, write $x=A(V)$ and $s=4\pi$. The expression equals $|xA(D\setminus V)-(s-x)A(D\cap V)|$, at most $x(s-x)$. A constant-version spherical isoperimetric inequality therefore gives the displayed bound, uniformly in nonempty open $D$. A smooth positive area density $\theta$ gives a second metric; integrating this bound over $D_t=\{\theta >t\}$ by the layer cake representation bounds the discrepancy between the two normalized area fractions by $\kappa (\max \theta -\min \theta )L^2/(s\tilde{A}(S))$.