Consider the n-dimensional cube ![{\displaystyle [0,1]^{n}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/40160923273b7109968df994dca832b91d957bf2)
with a Riemannian metric 
. Let
denote the distance between opposite faces of the cube. The Besicovitch inequality asserts that
![{\displaystyle \prod _{i}d_{i}\leq Vol([0,1]^{n},g)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fd34910c6caecad3df9c481007232314618b8902)
The inequality can be generalized in the following way. Given an n-dimensional Riemannian manifold M with connected boundary and a smooth map ![{\displaystyle f:M\rightarrow [0,1]^{n}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/93c1750d417f7a83802793ecee97072434f43806)
, such that the restriction of f to the boundary of M is a degree 1 map onto ![{\displaystyle \partial [0,1]^{n}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/94db59861d7f3c3ebeaa4bc13794eb1093038903)
, define
Then 
.
The Besicovitch inequality was used to prove systolic inequalities
on surfaces.[2][3]