Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 208 / 2 / d / Solution

The sharp Poincaré inequality for the uniform distribution on an interval isTensorizing these $n$ identical one-dimensional inequalities givesfor the uniform distribution on $[0,1{]}^n$. Therefore one may take $c=1/\pi$; this value is sharp, as functions depending only on one coordinate attain the one-dimensional constant.