[proofplan] We compare $K$ with its outer parallel bodies $K + r\overline{B}(0,1)$ as $r \downarrow 0$. The [Brunn-Minkowski inequality](/theorems/4080) gives a lower bound for the volume growth of these parallel bodies, while the [first variation formula](/theorems/2728) for convex bodies identifies the one-sided derivative of this volume at $r=0$ with the surface area $S(K)$. Passing to the limit yields the sharp lower bound for $S(K)$, and direct computation verifies equality for Euclidean balls. [/proofplan]

[step:Apply Brunn-Minkowski to the outer parallel body]Let \begin{align*} C := \overline{B}(0,1) \subset \mathbb{R}^n \end{align*} denote the closed Euclidean unit ball. Throughout the proof, let $\mathcal{L}^n$ denote $n$-dimensional Lebesgue measure on $\mathbb{R}^n$, and let $\mathcal{H}^{n-1}$ denote $(n-1)$-dimensional [Hausdorff measure](/page/Hausdorff%20Measure) on $\mathbb{R}^n$. For each $r > 0$ define the closed outer parallel body \begin{align*} K_r := K + rC = \{x + ry \in \mathbb{R}^n : x \in K,\ y \in C\}. \end{align*} Since $K$ is a convex body, $K$ is non-empty, compact, convex, and has non-empty interior. Hence $\mathcal{L}^n(K) > 0$, and $\mathcal{L}^n(C) > 0$. By the [Brunn-Minkowski inequality](/theorems/4114), applied to the measurable non-empty sets $K$ and $rC$, we obtain \begin{align*} \mathcal{L}^n(K_r)^{1/n} &\geq \mathcal{L}^n(K)^{1/n} + \mathcal{L}^n(rC)^{1/n}. \end{align*} By scaling of Lebesgue measure under the dilation $y \mapsto ry$, \begin{align*} \mathcal{L}^n(rC) = r^n \mathcal{L}^n(C), \end{align*} so \begin{align*} \mathcal{L}^n(K_r)^{1/n} &\geq \mathcal{L}^n(K)^{1/n} + r\mathcal{L}^n(C)^{1/n}. \end{align*}[/step]

[guided]The goal is to convert a global volume inequality into a surface area inequality. For a convex body, surface area measures the first-order growth of volume under outward thickening, so we thicken $K$ by a small Euclidean ball. Let \begin{align*} C := \overline{B}(0,1) \subset \mathbb{R}^n \end{align*} be the closed Euclidean unit ball. Throughout the proof, $\mathcal{L}^n$ denotes $n$-dimensional Lebesgue measure on $\mathbb{R}^n$, and $\mathcal{H}^{n-1}$ denotes $(n-1)$-dimensional Hausdorff measure on $\mathbb{R}^n$. For $r > 0$, define \begin{align*} K_r := K + rC = \{x + ry \in \mathbb{R}^n : x \in K,\ y \in C\}. \end{align*} This is the closed $r$-parallel body of $K$, so it is the correct object for the parallel-volume first variation formula used below. We apply the Brunn-[Minkowski inequality](/theorems/517) to the two measurable non-empty sets $K$ and $rC$. The hypotheses are satisfied because $K$ is a convex body, hence compact, Borel measurable, and non-empty, and $rC$ is a compact Euclidean ball, hence Borel measurable and non-empty. The inequality gives \begin{align*} \mathcal{L}^n(K + rC)^{1/n} &\geq \mathcal{L}^n(K)^{1/n} + \mathcal{L}^n(rC)^{1/n}. \end{align*} The dilation map \begin{align*} D_r: C &\to rC \\ y &\mapsto ry \end{align*} has Jacobian determinant $r^n$, so the change-of-variables formula for Lebesgue measure gives \begin{align*} \mathcal{L}^n(rC) = r^n\mathcal{L}^n(C). \end{align*} Taking $n$-th roots, with $r > 0$, yields \begin{align*} \mathcal{L}^n(rC)^{1/n} = r\mathcal{L}^n(C)^{1/n}. \end{align*} Therefore \begin{align*} \mathcal{L}^n(K_r)^{1/n} &\geq \mathcal{L}^n(K)^{1/n} + r\mathcal{L}^n(C)^{1/n}. \end{align*}[/guided]

[step:Convert the Brunn-Minkowski bound into a lower bound for volume growth]Define the positive constants \begin{align*} a &:= \mathcal{L}^n(K)^{1/n}, & b &:= \mathcal{L}^n(C)^{1/n}. \end{align*} The previous step gives \begin{align*} \mathcal{L}^n(K_r) \geq (a + rb)^n. \end{align*} Subtracting $\mathcal{L}^n(K) = a^n$ and dividing by $r > 0$, we obtain \begin{align*} \frac{\mathcal{L}^n(K_r) - \mathcal{L}^n(K)}{r} &\geq \frac{(a + rb)^n - a^n}{r}. \end{align*} Since the function \begin{align*} \phi: [0,\infty) &\to \mathbb{R} \\ t &\mapsto t^n \end{align*} is differentiable at $a$, we have \begin{align*} \lim_{r \downarrow 0}\frac{(a + rb)^n - a^n}{r} = \phi'(a)b = n a^{n-1} b. \end{align*} Thus \begin{align*} \liminf_{r \downarrow 0} \frac{\mathcal{L}^n(K_r) - \mathcal{L}^n(K)}{r} \geq n\mathcal{L}^n(K)^{(n-1)/n}\mathcal{L}^n(C)^{1/n}. \end{align*}[/step]

[guided]The previous step gives the lower bound \begin{align*} \mathcal{L}^n(K_r) \geq (a + rb)^n \end{align*} for every $r > 0$, where \begin{align*} a &:= \mathcal{L}^n(K)^{1/n}, & b &:= \mathcal{L}^n(C)^{1/n}. \end{align*} Both constants are positive: $K$ has non-empty interior because it is a convex body, and $C$ contains a non-empty open ball. Subtracting $\mathcal{L}^n(K) = a^n$ and dividing by the positive number $r$ preserves the inequality: \begin{align*} \frac{\mathcal{L}^n(K_r) - \mathcal{L}^n(K)}{r} &\geq \frac{(a + rb)^n - a^n}{r}. \end{align*} The right-hand side is the directional difference quotient for the differentiable function \begin{align*} \phi: [0,\infty) &\to \mathbb{R} \\ t &\mapsto t^n \end{align*} at the point $a$ in the direction $b$. Since $\phi'(a) = n a^{n-1}$, we obtain \begin{align*} \lim_{r \downarrow 0}\frac{(a + rb)^n - a^n}{r} = n a^{n-1}b. \end{align*} Taking the lower limit on the left-hand side therefore gives \begin{align*} \liminf_{r \downarrow 0} \frac{\mathcal{L}^n(K_r) - \mathcal{L}^n(K)}{r} \geq n\mathcal{L}^n(K)^{(n-1)/n}\mathcal{L}^n(C)^{1/n}. \end{align*}[/guided]

[step:Identify the first variation of volume with surface area]We use the first variation formula for parallel volume of convex bodies, in the following precise form: for a convex body $K \subset \mathbb{R}^n$ and the closed unit ball $C = \overline{B}(0,1)$, \begin{align*} S(K) = \mathcal{H}^{n-1}(\partial K) = \lim_{r \downarrow 0} \frac{\mathcal{L}^n(K + rC) - \mathcal{L}^n(K)}{r}. \end{align*} The hypotheses are satisfied because $K$ is a convex body. Applying this formula to the lower bound from the previous step gives \begin{align*} S(K) &\geq n\mathcal{L}^n(K)^{(n-1)/n}\mathcal{L}^n(C)^{1/n}. \end{align*}[/step]

[guided]The preceding step produced a lower bound for the infinitesimal growth rate of the volume of $K_r$. We now identify that growth rate with surface area. For a convex body $K \subset \mathbb{R}^n$, the first variation formula for parallel volume of convex bodies says that, with $C = \overline{B}(0,1)$, \begin{align*} S(K) = \mathcal{H}^{n-1}(\partial K) = \lim_{r \downarrow 0} \frac{\mathcal{L}^n(K + rC) - \mathcal{L}^n(K)}{r}. \end{align*} The theorem applies because $K$ is a convex body. This is the point where convexity supplies the needed regularity: the one-sided derivative of the closed parallel volume exists at $r=0$, and that derivative is the Hausdorff surface measure of the boundary. From the previous step, \begin{align*} \liminf_{r \downarrow 0} \frac{\mathcal{L}^n(K + rC) - \mathcal{L}^n(K)}{r} \geq n\mathcal{L}^n(K)^{(n-1)/n}\mathcal{L}^n(C)^{1/n}. \end{align*} Since the first variation formula gives the actual limit and identifies it with $S(K)$, we conclude \begin{align*} S(K) &\geq n\mathcal{L}^n(K)^{(n-1)/n}\mathcal{L}^n(C)^{1/n}. \end{align*}[/guided]

[step:Raise the surface area estimate to the $n$-th power]Both sides of \begin{align*} S(K) &\geq n\mathcal{L}^n(K)^{(n-1)/n}\mathcal{L}^n(C)^{1/n} \end{align*} are non-negative. Raising to the $n$-th power preserves the inequality and gives \begin{align*} S(K)^n &\geq n^n \mathcal{L}^n(K)^{n-1}\mathcal{L}^n(C). \end{align*} Since $C = \overline{B}(0,1)$ and $\mathcal{L}^n(\partial B(0,1)) = 0$, we have $\mathcal{L}^n(C) = \mathcal{L}^n(B(0,1))$. Hence \begin{align*} S(K)^n &\geq n^n \mathcal{L}^n(B(0,1)) \mathcal{L}^n(K)^{n-1}. \end{align*}[/step]

[guided]We start from the estimate proved in the previous step: \begin{align*} S(K) &\geq n\mathcal{L}^n(K)^{(n-1)/n}\mathcal{L}^n(C)^{1/n}. \end{align*} Both sides are non-negative because surface area and Lebesgue measure are non-negative. Therefore the map \begin{align*} \psi: [0,\infty) &\to [0,\infty) \\ t &\mapsto t^n \end{align*} is increasing on the relevant domain, so raising both sides to the $n$-th power preserves the inequality: \begin{align*} S(K)^n &\geq n^n \mathcal{L}^n(K)^{n-1}\mathcal{L}^n(C). \end{align*} Finally, $C = \overline{B}(0,1)$ differs from the open unit ball $B(0,1)$ only by the sphere $\partial B(0,1)$, and $\mathcal{L}^n(\partial B(0,1)) = 0$. Thus $\mathcal{L}^n(C) = \mathcal{L}^n(B(0,1))$, and the estimate becomes \begin{align*} S(K)^n &\geq n^n \mathcal{L}^n(B(0,1)) \mathcal{L}^n(K)^{n-1}. \end{align*}[/guided]

[step:Verify equality for Euclidean balls]Let $x_0 \in \mathbb{R}^n$ and $R > 0$, and set \begin{align*} K := \overline{B}(x_0,R). \end{align*} Then $K$ is a convex body. By translation invariance and scaling of Lebesgue measure, and because $\mathcal{L}^n(\partial B(x_0,R)) = 0$, \begin{align*} \mathcal{L}^n(K) = R^n \mathcal{L}^n(B(0,1)). \end{align*} Since $K + rC = \overline{B}(x_0,R+r)$ for every $r > 0$, the first variation formula gives \begin{align*} S(K) &= \lim_{r \downarrow 0} \frac{\mathcal{L}^n(\overline{B}(x_0,R+r)) - \mathcal{L}^n(\overline{B}(x_0,R))}{r} \\ &= \lim_{r \downarrow 0} \frac{(R+r)^n\mathcal{L}^n(B(0,1)) - R^n\mathcal{L}^n(B(0,1))}{r} \\ &= nR^{n-1}\mathcal{L}^n(B(0,1)). \end{align*} Therefore \begin{align*} S(K)^n &= \left(nR^{n-1}\mathcal{L}^n(B(0,1))\right)^n \\ &= n^n R^{n(n-1)}\mathcal{L}^n(B(0,1))^n, \end{align*} while \begin{align*} n^n \mathcal{L}^n(B(0,1))\mathcal{L}^n(K)^{n-1} &= n^n \mathcal{L}^n(B(0,1)) \left(R^n\mathcal{L}^n(B(0,1))\right)^{n-1} \\ &= n^n R^{n(n-1)}\mathcal{L}^n(B(0,1))^n. \end{align*} The two sides agree, so equality holds for closed Euclidean balls.[/step]

[guided]To verify the equality statement inside the class of convex bodies, we must use closed balls. Let $x_0 \in \mathbb{R}^n$ and $R > 0$, and define \begin{align*} K := \overline{B}(x_0,R). \end{align*} This set is non-empty, compact, convex, and has non-empty interior, hence it is a convex body. Translation invariance of Lebesgue measure and scaling under the dilation $y \mapsto x_0 + Ry$ give \begin{align*} \mathcal{L}^n(K) = R^n \mathcal{L}^n(B(0,1)), \end{align*} because the boundary sphere has $\mathcal{L}^n$-measure zero. The surface area can be computed from the same parallel-volume first variation used above. Since $K + rC = \overline{B}(x_0,R+r)$ for every $r > 0$, the first variation formula gives \begin{align*} S(K) &= \lim_{r \downarrow 0} \frac{\mathcal{L}^n(\overline{B}(x_0,R+r)) - \mathcal{L}^n(\overline{B}(x_0,R))}{r} \\ &= \lim_{r \downarrow 0} \frac{(R+r)^n\mathcal{L}^n(B(0,1)) - R^n\mathcal{L}^n(B(0,1))}{r} \\ &= nR^{n-1}\mathcal{L}^n(B(0,1)). \end{align*} Substituting this surface area into the left-hand side gives \begin{align*} S(K)^n &= \left(nR^{n-1}\mathcal{L}^n(B(0,1))\right)^n \\ &= n^n R^{n(n-1)}\mathcal{L}^n(B(0,1))^n. \end{align*} Substituting the volume into the right-hand side gives \begin{align*} n^n \mathcal{L}^n(B(0,1))\mathcal{L}^n(K)^{n-1} &= n^n \mathcal{L}^n(B(0,1)) \left(R^n\mathcal{L}^n(B(0,1))\right)^{n-1} \\ &= n^n R^{n(n-1)}\mathcal{L}^n(B(0,1))^n. \end{align*} The two quantities are identical, so equality holds for closed Euclidean balls.[/guided]