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  "body": "\n## Statement\n\nLet $n \\ge 2$, and let $\\omega_n$ be the volume of the unit ball of $\\R^n$.\nLet $\\Omega \\subseteq \\R^n$ be a bounded open set with $C^1$ boundary.\nLet $\\abs{\\Omega}$ be its [[Lebesgue measure|2Z2QVWZ262]] and let $\\abs{\\partial \\Omega}$ be the [[surface area|0S29Y4ZW0Y]] of its boundary.\nThen\n$$\\abs{\\partial \\Omega}^n \\ge n^n \\omega_n \\abs{\\Omega}^{n-1},$$\nand equality holds if and only if $\\Omega$ is a ball.\n\n## Remarks\n\nThe inequality follows from the [[Brunn–Minkowski theorem|9B02RPN7XD]].\nLet $B$ be the closed unit ball and $\\bar\\Omega$ the closure of $\\Omega$.\nThen $\\abs{\\bar\\Omega + \\varepsilon B} \\ge (\\abs{\\Omega}^{1/n} + \\varepsilon \\omega_n^{1/n})^n$.\nFor a $C^1$ boundary, $\\abs{\\partial \\Omega}$ is the limit of $(\\abs{\\bar\\Omega + \\varepsilon B} - \\abs{\\Omega}) / \\varepsilon$ as $\\varepsilon \\to 0$, and this limit is at least $n \\omega_n^{1/n} \\abs{\\Omega}^{(n-1)/n}$.\nRaising to the power $n$ gives the inequality; the equality case needs a separate argument.\nFor $n = 2$ it reads $L^2 \\ge 4 \\pi A$ for a region of area $A$ bounded by curves of total length $L$.\nExample: a square of side $1$ has $L = 4$ and $A = 1$, and indeed $16 \\ge 4\\pi$.\n\n## Explanations\n\n### L1 human\n\nAmong all shapes with a given volume, the ball has the smallest surface.\nEquivalently, among all shapes with a given surface area, the ball holds the most volume.\nThis is why soap bubbles are round: surface tension makes the surface as small as possible.\n\n### L2 human\n\nThe inequality has a version for sets of finite perimeter, which is the natural setting in geometric measure theory.\nIt is equivalent to the sharp Sobolev inequality $\\norm{u}_{n/(n-1)} \\le C_n \\norm{\\nabla u}_1$ for smooth compactly supported $u$, through the coarea formula.\nIn the plane it is often proved with Fourier series or by Steiner symmetrisation.\n",
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