Area of intersection between sphere and ball is less than surface area of ball Let $C=\{x\in \mathbb{R}^n: \|x-x_c\|_2 =r_c\}$ and $S = \{x\in \mathbb{R}^n: \|x-x_s\|_2 \le r_s\}$.
Suppose $n\ge 3$ and prove 
$$
\operatorname{Area}(C \cap S) \le \operatorname{Area}(\partial S), \forall x_c,x_s\in \mathbb{R}^n , \forall r_s,r_c\in \mathbb{R}_+
$$
where $\partial S$ is the boundary of $S$.
Intuitively, I believe this is true but I don't know how to prove it. Any hints may help, thank you.
 A: Since $C \cap S$ is part of the boundary of its convex hull $\verb/co/(C \cap S)$, we have
$$\verb/area/(C \cap S) \le \verb/area/(\partial(\verb/co/(C \cap S)))$$
We are going to show
$$\verb/area/(\partial(\verb/co/(C \cap S)) \le \verb/area/(\partial S)$$
Combine these two inequalities and we are done.

Let $V_n$ be the volume of the unit ball in $\mathbb{R}^n$ and
$\Omega_{n-1} = nV_n$ be the area of its boundary $S^{n-1}$.
For any unit vector $u \in S^{n-1}$, let $\pi_u : \mathbb{R}^n \to P_u$ be the
orthogonal projection of $\mathbb{R}^n$ onto a hyperplane $P_u$ having $u$ as normal vector. 
For any convex body $K \subset \mathbb{R}^n$ with piecewise smooth boundary,
let $A(K,u)$ be the area of its image under $\pi_u$ and $A(K)$ be the angular average over $S^{n-1}$.
$$A(K,u) \stackrel{def}{=} \verb/area/(\pi_u(K))
\quad\text{ and }\quad
A(K) \stackrel{def}{=} \frac{1}{\Omega_{n-1}}\int_{S^{n-1}} A(K,u)
$$
It is not hard to show following average projected area theorem:
$$A(K) = \frac{V_{n-1}}{\Omega_{n-1}} \verb/area/(\partial K)$$
As a corollary of this, if $K_1 \subset K_2$ are two convex bodies in $\mathbb{R}^n$, we have
$$A(K_1,u) \le A(K_2,u), \forall u \in S^{n-1}
\quad\implies\quad
\verb/area/(\partial K_1) \le \verb/area/(\partial K_2)$$
In particular, if we take $K_1 = \verb/co/(C\cap S)$ and $K_2 = S$, we obtain
the second inequality we need:
$$\verb/area/(\partial(\verb/co/(C \cap S)) \le \verb/area/(\partial S)$$
