Let $f: S^{n-1}\to \mathbb{R}$ be a continuous function ($S^{n-1}\subset \mathbb{R}^n$ is the unit sphere), $f(a)>0$ and $f(b)<0$ for certain points $a,b\in S^{n-1}$. By continuity these inequalities hold in $B_r(a)\cap S^{n-1}$ and $B_r(b)\cap S^{n-1}$ for a small radius $r$. Let $v:=\mathcal{H}^{n-2}(\partial (B_r(a)\cap S^{n-1}))=\mathcal{H}^{n-2}(\partial (B_r(b)\cap S^{n-1}))$ (taking the boundary with respect to the $S^{n-1}$-topology; the $n-2$-dimensional Hausdorff measure $v$ is roughly the surface area of the $n-2$-sphere of radius $r$ for small $r$).

Is it true that $\mathcal{H}^{n-2}(\{f=0\}) \ge v$, and how would one prove this? It seems geometrically obvious; the picture I have in mind looks as follows: Imagine the 2-sphere and two points $a,b$ on it. The zero set of $f$ is at least a curve which surrounds both $a$ and $b$, and the least length for such a curve is $v$, which is the length of the boundary of the neighbourhoods of $a$ and $b$ where $f$ is known to be positive/negative.