Geodesic
Refinement of Isoperimetric Inequality of Minkowski with the Account of Singularities in Boundaries of Intrinsic Parallel Bodies
Žurnal matematičeskoj fiziki, analiza, geometrii, Tome 10 (2014) no. 3, pp. 309-319 
V. I. Diskant. Refinement of Isoperimetric Inequality of Minkowski with the Account of Singularities in Boundaries of Intrinsic Parallel Bodies. Žurnal matematičeskoj fiziki, analiza, geometrii, Tome 10 (2014) no. 3, pp. 309-319. http://geodesic.mathdoc.fr/item/JMAG_2014_10_3_a2/
@article{JMAG_2014_10_3_a2,
     author = {V. I. Diskant},
     title = {Refinement of {Isoperimetric} {Inequality} of {Minkowski} with the {Account} of {Singularities} in {Boundaries} of {Intrinsic} {Parallel} {Bodies}},
     journal = {\v{Z}urnal matemati\v{c}eskoj fiziki, analiza, geometrii},
     pages = {309--319},
     year = {2014},
     volume = {10},
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     language = {en},
     url = {http://geodesic.mathdoc.fr/item/JMAG_2014_10_3_a2/}
}
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Voir la notice de l'article provenant de la source Math-Net.Ru

The following inequalities are proved: \begin{eqnarray*} S^n(A,B)\geq n^n\sum\limits_{i=0}^{k-1} V(B_{A_i})\left( V^{n-1}(A_i) - V^{n-1}(A_{i+1}) \right) +S^n(A_{-T}(B),B), \end{eqnarray*} \begin{eqnarray*} S^n(A,B)\geq n^n\int\limits_{0}^{T} g(t) df(t) +S^n(A_{-T}(B),B), \end{eqnarray*} \begin{eqnarray*} S^n(A,B)\geq n^n\int\limits_{0}^{q} g(t) df(t) +S^n(A_{-q}(B),B), \end{eqnarray*} where $V(A)$, $V(B)$ stand for the volumes of convex bodies $A$ and $B$ in $\mathbb R^n$ ($n\geq 2$), $S(A,B)$ denotes the area of the surface of the body $A$ relative to the body $B$, $q$ is the capacity factor of the body $B$ with respect to the body $A$, $A_i = A_{-t_i}(B) = A / (t_iB)$ is the inner body parallel to the body $A$ with respect to the body $B$ at a distance $t_i$, $0=t_0 < t_1 <\ldots< t_i< \ldots < t_{k-1}, $B_{A_i}$ is a shape body of $A_i$ relative to $B$, $g(t) = V(B_{A_{-t}(B)})$, $f(t) = - V^{n-1}( A_{-t}(B))$, $\int\limits_{0}^{T} g(t) df(t) $ is the Riemann–Stieltjes integral of the function $g(t)$ by the function $f(t)$, and $\int\limits_{0}^{q} g(t) df(t) = \lim\limits_{T\to q} \int\limits_{0}^{T} g(t) df(t)$.

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