Surface area of a convex set less than that of its enclosing sphere?

Is the boundary measure ("surface area") of a convex set in $\mathbb{R}^n$ less than the boundary measure of it's enclosing hypersphere (smallest hypersphere that contains the set)?

In 2D I've found a paper that states it's true, and intuitively I think it should be true for n-D, but I'm having trouble proving it.

Edit Possible solution: approximate the boundary of the inner set with panels - line segments/triangles/tetrahedra/ect depending on the dimension. Then orthogonally project those panels onto the hypersphere. Since the set is convex, the projections of the panels don't overlap, and since the projections are orthogonal, their projections onto the sphere are larger than the original panels.

Edit 2 More general conjecture: If $X$ and $Y$ are convex sets with $X \subset Y \subset \mathbb{R}^n$, then $|\partial X| \le |\partial Y|$

Let the set be $C$ and the sphere $S$. Map $S$ onto $C$ so that $f(s)$ is the closest point to $s$ in $C$. Then $f$ is a contraction, so it decreases Hausdorff measure of any dimension.

To see that $f$ is a contraction, let $f(s_1) = c_1$ and $f(s_2) = c_2 \ne c_1$. Then $(c_1 - c_2) \cdot (s_1 - c_1) \ge 0$ (otherwise you get a closer point to $s_1$ by moving from $c_1$ a short distance in the direction of $c_2$), and similarly $(c_2 - c_1) \cdot (s_2 - c_2) \ge 0$. Add these and rearrange to get $\|c_1 - c_2\|^2 \le (c_1 - c_2) \cdot (s_1 - s_2)$, and use the Cauchy-Schwarz inequality.