1981 年, 杨路, 张景中 ^[1] 两位教授率先将三角形中的 Pedoe 不等式推广到 n 维欧氏空间的单形后, 涉及两个单形的不等式的研究备受关注 ^{[2- 4]}, 但联系多个单形的几何关系的文献并不多见. 最近, 笔者在文献 [5] 中, 给出了涉及一个单形的棱长与另一个单形的内点, 高,n - 1 维侧面的体积, 外接超球半径的几个不等式. 本文中, 我们用距离几何的理论和方法,结合杨 - 张不等式的推广式 ^[6], 建立了涉及四个单形的一类不等式. 即两个单形的棱长与另两个单形的内点, 中线, 高, 外接超球半径, 内切超球半径, 旁切超球半径以及 n - 1 维侧面的体积, 外接超球半径, 内切超球半径的一类新的几何不等式. 特别地, 获得了联系三个单形以及两个单形的一些新的几何不等式, 推广了文献 [5] 中的所有结果. 最后, 提出了有待进一步讨论的一个猜想.

全文约定: 用 $\Omega_{k}=A_{k1}A_{k2}\cdots A_{kn+1}$ $(k=1,2,3,4)$ 表示 $n$ 维欧氏空间 $R^{n}$ 中的四个 $n$ 维单形, 其棱长为 $\left\vert A_{ki} A_{kj}\right\vert =\rho_{kij}$ $(k=1,2,3,4;$ $1\leq i<j\leq n+1),$ 其体积, 外接超球半径, 内切超球半径分别为 $ V_{k},R_{k},r_{k}$ $\ (k=1,2,3,4),$ 顶点 $A_{ki}$ 所对的侧面 $f_{ki}=\Omega{ A_{ki}} $ 的旁切超球半径为 $r_{ki}^{\prime},$ 侧面 $ f_{ki}$ 的 $n-1$ 维体积为 $S_{ki},$ 侧面 $f_{ki}$ 上的高为 $h_{ki},$ 侧面 $f_{ki}$ 上的中线长为 $l_{ki},$ 侧面 $f_{ki}$ 的外接超球半径为 $R_{ki},$ 内切超球半径为 $r_{ki},$ $\Omega_{k}$ 内任意一点 $P_{k}$ 到该面的距离为 $ d_{ki} (k=1,2,3,4;$ $i=1,2,3,\cdots,n+1).$ 对一个 $n$ 维单形 $\Omega=A_{1} A_{2}\cdots A_{n+1},$ 设其体积, 外接超球半径, 内切超球半径分别为 $V,R,r$, 顶点 $A_{i}$ 所对的侧面 $f_{i}=\Omega{ A_{i}} $ 的旁切超球半径为 $r^{\prime}$, 侧面 $\ f_{i}$ 的 $n-1$ 维体积为 $S_{i},$ 侧面 $f_{i}$ 上的高为 $h_{i},$ 侧面 $ f_{i}$ 上的中线长为 $l_{i},$ 侧面 $f_{i}$ 的外接超球半径为 $R_{i}$, 侧面 $f_{i}$ 的内切超球半径为 $r_{i},$ $\Omega$ 内任意一点 $P$ 到该面的距离为 $d_{i} (i=1,2,\cdots ,n+1).$

引理 2.1^[6]用 $\Phi=\left\{ A_{i} (m_{i}),i=1,2,\cdots,N\right\} $ 和 $\Phi^{\prime}=\left\{ A_{i}^{\prime }(m_{i}^{\prime}),i=1,2,\cdots,N\right\} $ $(N>n)$ 分别表示 $n$ 维欧式空间 $R^{n}$ 中具有相同质点数量的, 并且是完全同向的两个质点组, $m_{i} >0,m_{i}^{\prime}>0$ 是点 $A_{i}$ 与 $A_{i}^{\prime}$ 所赋有的质量, 分别取 $\Phi$ 和 $\Phi^{\prime}$ 的 $k+1$ 个点 $A_{i_{0}},A_{i_{1}} ,\cdots,A_{i_{k}}$ 和 $A_{i_{0}}^{\prime},A_{i_{1}}^{\prime},\cdots,A_{i_{k} }^{\prime}$ 所支撑的单形的 $k$ 维有向体积记为 $\ V_{i_{0}i_{1}\cdots i_{k}}$ 和 $V_{i_{1}i_{2}\cdots i_{k}}^{\prime}$.

令

$W_{0} =\sum\limits_{i=1}^{N}\sqrt{m_{i}m_{i}^{\prime}},\\ W_{k} =\underset{i_{0}<i_{1}<\cdots<i_{k}}{\sum\sum\cdots\sum} \sqrt{m_{i_{0}}m_{i_{0}}^{\prime}m_{i_{1}}m_{i_{1}}^{\prime}\cdots m_{i_{k} }m_{i_{k}}^{\prime}}V_{i_{0}i_{1}\cdots i_{k}}V_{i_{0}i_{1}\cdots i_{k} }^{\prime}(1\leq k\leq n),$

则有不等式

$\frac{W_{k}^{l}}{W_{l}^{k}}\geq\frac{\left[ (n-l)!(l!)^{3}\right] ^{k} }{\left[ (n-k)!(k!)^{3}\right] ^{l}}(n!W_{0})^{l-k}(1\leq k<l\leq n),$

(2.1)

当且仅当 $\Phi$ 与 $\Phi^{\prime}$ 的广义惯量椭球为球时, (2.1) 式取等号.

引理 2.2^{[ 7]}在$n$ 维单形 $\Omega$ 中, 有

$V\leq\sqrt{n+1}(\frac{n!^{2}}{n^{3n}})^{1/2(n-1)}\prod\limits_{i=1}^{n+1} S_{i}^{n/(n^{2}-1)}, $

(2.2)

当且仅当 $\Omega$ 为正则单形时, (2.2) 式取等号.

引理 2.3^[8]在 $n$ 维单形 $\Omega$ 中$,$ 有

$\prod\limits_{i=1}^{n+1}S_{i}\leq\left[ \frac{(n+1)^{n-1}}{(n-1)!^{2}n^{n-2} }\right] ^{(n+1)/2}R^{n^{2}-1},$

(2.3)

$V\leq\frac{1}{n!}\left[ \frac{n^{n}}{(n+1)^{n-1}}\right] ^{1/2} \prod\limits_{i=1}^{n+1}l_{i}^{n/(n+1)},$

(2.4)

当且仅当 $\Omega$ 为正则单形时, 以上两式取等号.

引理 2.4^[9]在 $n$ 维单形 $\Omega$ 中$,$ 有

$\sum\limits_{i=1}^{n+1}\frac{1}{r_{i}^{2}}\leq\frac{n-1}{r^{2}},$

(2.5)

且仅当 $\Omega$ 为正则单形时, (2.5) 式取等号.

引理 2.5 ^[10]在 $n$ 维单形 $\Omega$ 中, 对 $0<\theta\leq1,$ 有

$\sum\limits_{i=1}^{n+1}\prod\limits_{j=1,j\neq i}^{n+1}(h_{j}-d_{j})^{\theta }\leq(n+1)\left( \frac{n^{n}(n!)^{2}}{(n+1)^{n+1}}\right) ^{\theta /2}V^{\theta},$

(2.6)

$\sum\limits_{i=1}^{n+1}\prod\limits_{j=1,j\neq i}^{n+1}(h_{j}+d_{j})^{\theta }\leq(n+1)\left( \frac{(n+2)^{2n}(n!)^{2}}{(n+1)^{n+1}n^{n}}\right) ^{\theta/2}V^{\theta},$

(2.7)

当且仅当 $\Omega$ 为正则单形且 $P$ 为其中心时, 以上两式取等号.

定理 1 在 $n$ 维单形 $\Omega _{1},\Omega_{2},\Omega_{3},\Omega_{4}$ 中, 对 $\alpha,\beta>0,$ 有

$\sum\limits_{1 \le i<j \le n + 1}\frac{\rho_{1ij}\rho_{2ij}}{(d_{3i}d_{3j})^{\alpha }(d_{4i}d_{4j})^{\beta}}\geq(n!)^{2(1-\alpha-\beta)/n}n^{\alpha+\beta +1}(n+1)^{[(n+1)(\alpha+\beta)+n-1]/n}\left( \frac{V_{1}V_{2}}{V_{3} ^{2\alpha}V_{4}^{2\beta}}\right) ^{1/n},$

(3.1)

当且仅当 $\Omega_{1},\Omega_{2},\Omega_{3},\Omega_{4}$ 均为正则单形且 $P_{3}$ 与 $P_{4}$ 分别为 $\Omega_{3}$ 与 $\Omega_{4}$ 的中心时, (3.1) 式取等号.

证 在引理 2.1 中, 取 $l=n-1$, $k=1.$ 则有

${\left( {\sum\limits_{1 \le i<j \le n + 1} {} \sqrt {{m_i}{m_j}m_i^\prime m_j^\prime } {\rho _{1ij}}{\rho _{2ij}}} \right)^{n - 1}} \\ \ge n{!^2}{n^{n - 4}}{\left( {\sum\limits_{i = 1}^{n + 1} {\sqrt {{m_i}m_i^\prime } } } \right)^{n - 2}}\left( {\sum\limits_{i = 1}^{n + 1} {\prod\limits_{j = 1,j \ne i}^{n + 1} {\sqrt {{m_j}m_j^\prime } } } {S_{1i}}{S_{2i}}} \right),$

(3.2)

当且仅当 $\Omega_{1},\Omega_{2}$ 均为正则单形且 $m_{1}=m_{2} =\cdots=m_{n+1},m_{1}^{\prime}=m_{2}^{\prime}=\cdots=m_{n+1}^{\prime}$ 时, (3.2) 式取等号.

在不等式 (3.2) 中, 令 $m_{i}=(1/d_{3i})^{2\alpha},m_{i}^{\prime} =(1/d_{4i})^{2\beta}$ $(i=1,2,\cdots,n+1),$ 并运用平均值不等式, 同时注意到 $\sum\limits_{i=1}^{n+1}S_{i}d_{i}=nV,$ 得

$\left( \sum\limits_{1 \le i<j \le n + 1}\frac{\rho_{1ij}\rho_{2ij}}{(d_{3i} d_{3j})^{\alpha}(d_{4i}d_{4j})^{\beta}}\right) ^{n-1} \\ \geq n!^{2}n^{n-4}(n+1)^{n-1}\left( \prod\limits_{i=1}^{n+1}\frac {1}{d_{3i}^{\alpha}d_{4i}^{\beta}}\right) ^{2(n-1)/(n+1)}\prod\limits_{i=1} ^{n+1}\left( S_{1i}S_{2i}\right) ^{1/(n+1)} \\ =n!^{2}n^{n-4}(n+1)^{n-1}\left( \prod\limits_{i=1}^{n+1}\frac {S_{3i}^{\alpha}S_{4i}^{\beta}}{(d_{3i}S_{3i})^{\alpha}(d_{4i}S_{4i})^{\beta} }\right) ^{2(n-1)/(n+1)}\prod\limits_{i=1}^{n+1}(S_{1i}S_{2i})^{1/(n+1)}\\ \geq n!^{2}n^{n-4}(n+1)^{n-1}\left[ \frac{\prod\limits_{i=1}^{n+1} S_{3i}^{\alpha}\prod\limits_{i=1}^{n+1}S_{4i}^{\beta}}{\left( \sum \limits_{i=1}^{n+1}\dfrac{S_{3i}d_{3i}}{n+1}\right) ^{(n+1)\alpha}\left( \sum\limits_{i=1}^{n+1}\dfrac{S_{4i}d_{4i}}{n+1}\right) ^{(n+1)\beta} }\right] ^{\frac{2(n-1)}{n+1}}\prod\limits_{i=1}^{n+1}(S_{1i}S_{2i} )^{1/(n+1)}\\ =n!^{2}n^{n-4-2(n-1)(\alpha+\beta)}(n+1)^{(n-1)[2(\alpha+\beta)+1]}\left( \prod\limits_{i=1}^{n+1}S_{1i}S_{2i}S_{3i}^{2(n-1)\alpha}S_{4i}^{2(n-1)\beta }\right) ^{1/(n+1)}\left( \frac{1}{V_{3}^{\alpha}V_{4}^{\beta}}\right) ^{2(n-1)}. $

再将 (2.2) 式代入并整理, 可得

$\left( \sum\limits_{1 \le i<j \le n + 1}\frac{\rho_{1ij}\rho_{2ij}}{(d_{3i} d_{3j})^{\alpha}(d_{4i}d_{4j})^{\beta}}\right) ^{n-1}\\ \geq(n!)^{2(n-1)(1-\alpha-\beta)/n}n^{(n-1)(\alpha+\beta+1)} (n+1)^{(n-1)[(n+1)(\alpha+\beta)+n-1]/n}\left( \frac{V_{1}V_{2}} {V_{3}^{2\alpha}V_{4}^{2\beta}}\right) ^{(n-1)/n}.$

上式两边同开 $(n-1)$ 次方, 即得不等式 (3.1). 由证明过程可知, (3.1) 式取 等号的条件同定理 1 所述.

在定理 1 中, 取 $P_{3}$ 与 $P_{4}$ 分别为单形 $\Omega_{3}$ 与 $\Omega_{4}$ 的重心, 此时有 $d_{3i}=h_{3i}/(n+1),$ $d_{4i}=h_{4i}/(n+1)$ $(i=1,2,\cdots,n+1),$ 可得

推论 1 在 $n$ 维单形 $\ \Omega _{1},\Omega_{2},\Omega_{3},\Omega_{4}$ 中, 对 $\alpha,\beta>0,$ 有

$\sum\limits_{1 \le i<j \le n + 1}\frac{\rho_{1ij}\rho_{2ij}}{\left( h_{3i} h_{3j}\right) ^{\alpha}\left( h_{4i}h_{4j}\right) ^{\beta}}\geq (n!)^{2(1-\alpha-\beta)/n}n^{\alpha+\beta+1}(n+1)^{[(1-n)(\alpha +\beta)+n-1]/n}\left( \frac{V_{1}V_{2}}{V_{3}^{2\alpha}V_{4}^{2\beta} }\right) ^{1/n},$

(3.3)

当且仅当 $\Omega_{1},\Omega_{2},\Omega_{3},\Omega_{4}$ 均为正则单形时, (3.3) 式取等号.

若注意到 $h_{3i}=nV_{3}/S_{3i},$ $h_{4i}=nV_{4}/S_{4i}$ $(i=1,2,\cdots,n+1).$ 由 (3.3) 式, 可得

推论 2 在 $n$ 维单形 $\Omega _{1},\Omega_{2},\Omega_{3},\Omega_{4}$ 中, 对 $\alpha,\beta>0,$ 有

$\sum\limits_{1 \le i<j \le n + 1}\rho_{1ij}\rho_{2ij}\left( S_{3i}S_{3j}\right) ^{\alpha}(S_{4i}S_{4j})^{\beta}\nonumber\\ \geq(n!)^{2(1-\alpha-\beta)/n}n^{3(\alpha+\beta)+1}(n+1)^{[(1-n)(\alpha +\beta)+n-1]/n}\left( V_{1}V_{2}\right) ^{1/n}(V_{3}^{\alpha}V_{4}^{\beta })^{2(n-1)/n}.$

(3.4)

(3.4) 式取等号的条件同推论 1.

在定理 1 中, 取 $P$ 为单形 $\Omega$ 的外心 $O$, 则 $\left\vert OA_{i}\right\vert =R.$ 由于 $O$ 在各侧面 $f_{i}$ 上的射影是各侧面的外心 $O_{i}$ $(i=1,2,\cdots,n+1),$ 则

$\left\vert O_{i}A_{1}\right\vert =\cdots=\left\vert O_{i}A_{i-1}\right\vert =\left\vert O_{i}A_{i+1}\right\vert =\cdots=\left\vert O_{i}A_{n+1}\right\vert =R_{i}.$

故

$d_{i}^{2}=\left\vert O_{i}O\right\vert ^{2}=R^{2}-R_{i}^{2}(i=1,2,\cdots,n+1).$

于是有

推论 3 在$n$ 维单形 $\Omega _{1},\Omega_{2},\Omega_{3},\Omega_{4}$ 中, 对 $\alpha,\beta>0,$ 有

$\sum\limits_{1 \le i<j \le n + 1}\frac{\rho_{1ij}\rho_{2ij}}{[(R_{3}^{2}-R_{3i} ^{2})(R_{3}^{2}-R_{3j}^{2})]^{\alpha}[(R_{4}^{2}-R_{4i}^{2})(R_{4}^{2} -R_{4j}^{2})]^{\beta}}\nonumber\\ \geq(n!)^{2[1-2(\alpha+\beta)]/n}n^{2(\alpha+\beta)+1}(n+1)^{[2(n+1)(\alpha +\beta)+n-1]/n}\left( \frac{V_{1}V_{2}}{V_{3}^{4\alpha}V_{4}^{4\beta} }\right) ^{1/n}.$

(3.5)

(3.5) 式取等号的条件同推论 1.

定理 2 在$n$ 维单形 $\Omega _{1},\Omega_{2},\Omega_{3},\Omega_{4}$ 中, 对 $\alpha,\beta>0,$ 有

$\sum\limits_{1 \le i<j \le n + 1} {} \frac{{{\rho _{1ij}}{\rho _{2ij}}}}{{{{({S_{3i}}{S_{3j}})}^\alpha }{{({S_{4i}}{S_{4j}})}^\beta }}}\nonumber \\ \ge {(n!)^{(1 + \alpha + \beta )/n}}{n^{(n - 1)(\alpha + \beta ) + 1}}{(n + 1)^{(n - 1)[1 - n(\alpha + \beta )]/n}}{({V_1}{V_2})^{1/n}}{\left( {\frac{1}{{R_3^\alpha R_4^\beta }}} \right)^{2(n - 1)}},$

(3.6)

$\sum\limits_{1 \le i<j \le n + 1} \rho_{1ij}\rho_{2ij}(l_{3i}l_{3j})^{\alpha} (l_{4i}l_{4j})^{\beta}\nonumber\\ \geq(n!)^{2(1+\alpha+\beta)/n}n^{1-\alpha-\beta}(n+1)^{(n-1)(1+\alpha +\beta)/n}(n-1)^{-(\alpha+\beta)}(V_{1}V_{2}V_{3}^{2\alpha}V_{4}^{2\beta })^{1/n},$

(3.7)

$\sum\limits_{1 \le i<j \le n + 1} \rho_{1ij}\rho_{2ij}(r_{3i}r_{3j})^{\alpha} (r_{4i}r_{4j})^{\beta}\nonumber\\ \geq(n!)^{2/n}n(n+1)^{[n(1+\alpha+\beta)-1]/n}(n-1)^{-(\alpha+\beta)} (V_{1}V_{2})^{1/n}(r_{3}^{\alpha}r_{4}^{\beta})^{2},$

(3.8)

$\sum\limits_{1 \le i<j \le n + 1}\rho_{1ij}\rho_{2ij}(r_{3i}^{\prime}r_{3j}^{\prime })^{\alpha}(r_{4i}^{\prime}r_{4j}^{\prime})^{\beta}\nonumber\\ \geq(n!)^{2/n}n(n+1)^{[n+2n(\alpha+\beta)-1]/2}(n-1)^{-2(\alpha+\beta )}(V_{1}V_{2})^{1/n}(r_{3}^{\alpha}r_{4}^{\beta})^{2},$

(3.9)

当且仅当 $\Omega_{1},\Omega_{2},\Omega_{3},\Omega_{4}$ 均为正则单形时, (3.6)-(3.9) 四式取等号.

证 对不等式 (3.2) 右边运用平均值不等式, 得

$\left( \sum\limits_{1 \le i<j \le n + 1}\sqrt{m_{i}m_{j}m_{i}^{\prime}m_{j} ^{\prime}}a_{1ij}a_{2ij}\right) ^{n-1}\nonumber\\ \geq(n-1)!^{2}n^{n-2}(n+1)^{n-1}\left( \prod\limits_{i=1}^{n+1}m_{i} m_{i}^{\prime}\right) ^{(n-1)/(n+1)}\prod\limits_{i=1}^{n+1}\left( S_{1i}S_{2i}\right) ^{1/(n+1)}.$

(3.10)

在 (3.10)式中, 令 $m_{i}=(1/S_{3i})^{2\alpha},m_{i}^{\prime}=(1/S_{4i} )^{2\alpha}$$(i=1,2,\cdots,n+1),$ 并注意到不等式 (2.2), (2.3) 两式即 可证得不等式 (3.6).

在不等式 (3.10) 中, 令 ${m_i} = r_{3i}^{2\alpha },$ $m_{i}^{\prime} =l_{4i}^{2\alpha}$ $(i=1,2,\cdots,n+1),$ 并注意到不等式 (2.2), (2.4) 两式 即可证得不等式 (3.7).

在不等式 (2.5)中, 运用平均值不等式, 得

$\prod\limits_{i=1}^{n+1}r_{i}\geq\left( \frac{n+1}{n-1}\right) ^{(n+1)/2}r^{n+1}.$

在不等式 (3.10) 中, 令 $m_{i}=r_{3i}^{2\alpha},m_{i}^{\prime} =r_{4i}^{2\beta}$ $(i=1,2,\cdots,n+1),$ 由上式并注意到不等式 (2.2) 即可 证得不等式 (3.8).

对恒等式^[8] $\sum\limits_{i=1}^{n+1}\dfrac{1}{r^{\prime}} =\dfrac{n-1}{r}.$ 运用平均值不等式, 得

$\prod\limits_{i=1}^{n+1}r_{i}^{\prime}\geq\left( \frac{n+1}{n-1}\right) ^{n+1}r^{n+1}.$

在不等式 (3.10)中, 令 $m_{i}=r_{3i}^{\prime2\alpha},m_{i}^{\prime} =r_{4i}^{\prime2\beta}$ $(i=1,2,\cdots,n+1),$ 由上式和不等式 (2.2) 即可证 得不等式 (3.9). 且易知 (3.6)-(3.9) 四式取等号的条件为定理 2 所述. 定理 2 证毕.

定理 3 在$n$ 维单形 $\Omega _{1},\Omega_{2},\Omega_{3},\Omega_{4}$ 中, 对 $\alpha,\beta>0,$ 有

$\sum\limits_{1\leq i<j\leq n+1}\frac{\rho_{1ij}\rho_{2ij}}{[(h_{3i}-d_{3i} )(h_{3j}-d_{3j})]^{\alpha}[(h_{4i}-d_{4i})(h_{4j}-d_{4j})]^{\beta}}\nonumber\\ \geq(n!)^{2(1-\alpha-\beta)/n}n^{1-\alpha-\beta}(n+1)^{[(n+1)(\alpha +\beta)+n-1]/n}\left( \frac{V_{1}V_{2}}{V_{3}^{2\alpha}V_{4}^{2\beta} }\right) ^{1/n},$

(3.11)

$\sum\limits_{1\leq i<j\leq n+1}\frac{\rho_{1ij}\rho_{2ij}}{[(h_{3i}+d_{3i} )(h_{3j}+d_{3j})]^{\alpha}[(h_{4i}+d_{4i})(h_{4j}+d_{4j})]^{\beta}}\nonumber\\ \geq(n!)^{2(1-\alpha-\beta)/n}n^{1+\alpha+\beta}(n+1)^{[(n+1)(\alpha +\beta)+n-1]/n}(n+2)^{-2(\alpha+\beta)}\left( \frac{V_{1}V_{2}} {V_{3}^{2\alpha}V_{4}^{2\beta}}\right) ^{1/n}.$

(3.12)

(3.11), (3.12) 两式取等号的条件同定理 1.

证 我们仅证 (3.11) 式, 而 (3.12) 式类似证之.

在不等式 (2.6) 中, 取 $\theta=1,$ 并对其左边运用算术-几何平均值不等式, 得

$\prod\limits_{i=1}^{n+1}(h_{i}-d_{i})^{n/(n+1)}\leq\left( \frac{n^{n} (n!)^{2}}{(n+1)^{n+1}}\right) ^{1/2}V.$

(3.13)

在不等式 (3.10) 中, 取 $m_{i}=1/(h_{3i}-d_{3i})^{2\alpha},m_{i}^{\prime }=1/(h_{4i}-d_{4i})^{2\beta}$ $(i=1,2,\cdots,n+1),$ 并注意到不等式 (2.2), (3.13) 两式即可证得不等式 (3.11), 且易知 (3.11) 式取等号的条件为定理 3 所述.

在定理 1, 2, 3 和推论 1, 2, 3中, 取 $\Omega_{2}$ 为 $\Omega_{1},$ 可得 联系三个单形的一些新的不等式.

推论 4 在 $n$ 维单形 $\Omega_{1},\Omega _{2},\Omega_{3}$ 中, 对 $\alpha,\beta>0,$ 有

$\sum\limits_{1\leq i<j \leq n+1}\frac{\rho_{1ij}^{2}}{(d_{2i}d_{2j})^{\alpha} (d_{3i}d_{3j})^{\beta}}\nonumber\\ \geq(n!)^{2(1-\alpha-\beta)/n}n^{\alpha+\beta +1}(n+1)^{[(n+1)(\alpha+\beta)+n-1]/n}\left( \frac{V_{1}}{V_{2}^{\alpha} V_{3}^{\beta}}\right) ^{2/n},$

(3.14)

$\sum\limits_{1\leq i<j \leq n+1}\frac{\rho_{1ij}^{2}}{(h_{2i}h_{2j})^{\alpha} (h_{3i}h_{3j})^{\beta}}\nonumber\\ \geq (n!)^{2(1-\alpha-\beta)/n}n^{\alpha+\beta +1}(n+1)^{[(1-n)(\alpha+\beta)+n-1]/n}\left( \frac{V_{1}}{V_{2}^{\alpha} V_{3}^{\beta}}\right) ^{2/n},$

(3.15)

$\sum\limits_{1\leq i<j\leq n+1}\rho_{1ij}^{2}(S_{2i}S_{2j})^{\alpha}(S_{3i} S_{3j})^{\beta}\nonumber\\ \geq (n!)^{2(1-\alpha-\beta)/n}n^{3(\alpha+\beta)+1}(n+1)^{[(1-n)(\alpha +\beta)+n-1]/n}V_{1}^{2/n}(V_{2}^{\alpha}V_{3}^{\beta})^{2(n-1)/n},$

(3.16)

$\sum\limits_{1\leq i<j\leq n+1}\frac{\rho_{1ij}^{2}}{[(R_{2}^{2}-R_{2i}^{2} )(R_{2}^{2}-R_{2j}^{2})]^{\alpha}[(R_{3}^{2}-R_{3i}^{2})(R_{3}^{2}-R_{3j} ^{2})]^{\beta}}\nonumber\\ \geq (n!)^{2[1-2(\alpha+\beta)]/n}n^{2(\alpha+\beta)+1}(n+1)^{[2(n+1)(\alpha +\beta)+n-1]/n}\left( \frac{V_{1}}{V_{2}^{2\alpha}V_{3}^{2\beta}}\right) ^{2/n},$

(3.17)

$\sum\limits_{1\leq i<j\leq n+1}\frac{\rho_{1ij}^{2}}{(S_{2i}S_{2j})^{\alpha }(S_{3i}S_{3j})^{\beta}}\nonumber\\ \geq (n!)^{(1+\alpha+\beta)/n}n^{(n-1)(\alpha+\beta)+1} (n+1)^{(n-1)[1-n(\alpha+\beta)]/n}V_{1}^{2/n}\left( \frac{1}{R_{2}^{\alpha }R_{3}^{\beta}}\right) ^{2(n-1)},$

(3.18)

$\sum\limits_{1\leq i<j\leq n+1}\rho_{1ij}^{2}(l_{2i}l_{2j})^{\alpha}(l_{3i} l_{3j})^{\beta}\nonumber\\ \geq(n!)^{2(1+\alpha+\beta)/n}n^{1-\alpha-\beta}(n+1)^{(n-1)(1+\alpha +\beta)/n}(n-1)^{-(\alpha+\beta)}(V_{1}V_{2}^{\alpha}V_{3}^{\beta})^{2/n},$

(3.19)

$\sum\limits_{1\leq i<j\leq n+1}\rho_{1ij}^{2}(r_{2i}r_{2j})^{\alpha}(r_{3i} r_{3j})^{\beta}\nonumber\\ \geq(n!)^{2/n}n(n+1)^{[n(1+\alpha+\beta)-1]/n}(n-1)^{-(\alpha +\beta)}V_{1}^{2/n}(r_{2}^{\alpha}r_{3}^{\beta})^{2},$

(3.20)

$\sum\limits_{1\leq i<j\leq n+1}\rho_{1ij}^{2}(r_{2i}^{\prime}r_{2j}^{\prime} )^{\alpha}(r_{3i}^{\prime}r_{3j}^{\prime})^{\beta}\nonumber\\ \geq (n!)^{2/n} n(n+1)^{[n+2n(\alpha+\beta)-1]/2}(n-1)^{-2(\alpha+\beta)}V_{1}^{2/n} (r_{2}^{\alpha}r_{3}^{\beta})^{2},$

(3.21)

$\sum\limits_{1\leq i<j\leq n+1}\frac{\rho_{1ij}^{2}}{[(h_{2i}-d_{2i} )(h_{2j}-d_{2j})]^{\alpha}[(h_{3i}-d_{3i})(h_{3j}-d_{3j})]^{\beta}}\nonumber\\ \geq (n!)^{2(1-\alpha-\beta)/n}n^{1-\alpha-\beta}(n+1)^{[(n+1)(\alpha +\beta)+n-1]/n}\left( \frac{V_{1}}{V_{2}^{\alpha}V_{3}^{\beta}}\right) ^{2/n},$

(3.22)

$\sum\limits_{1\leq i<j\leq n+1}\frac{\rho_{1ij}^{2}}{[(h_{2i}+d_{2i} )(h_{2j}+d_{2j})]^{\alpha}[(h_{3i}+d_{3i})(h_{3j}+d_{3j})]^{\beta}}\nonumber\\ \geq (n!)^{2(1-\alpha-\beta)/n}n^{1+\alpha+\beta}(n+1)^{[(n+1)(\alpha +\beta)+n-1]/n}(n+2)^{-2(\alpha+\beta)}\left( \frac{V_{1}}{V_{2}^{\alpha }V_{2}^{\beta}}\right) ^{2/n},$

(3.23)

当且仅当单形 $\Omega_{1},\Omega_{2},\Omega_{3}$ 均为正则单形且 $P_{2}$ 与 $ P_{3}$ 分别为 $\Omega_{2}$ 与 $\Omega_{3}$ 的中心时, (3.14), (3.22), (3.23) 式取等号, 当且仅当单形 $ \Omega_{1},\Omega_{2},\Omega_{3}$ 均为正 则单形时, (3.15)-(3.21) 式取等号.

在定理 1, 2, 3 和推论 1, 2, 3 中, 取 $ \Omega_{4}$ 为 $\Omega_{3},$ 令 $\alpha$ 为 $\dfrac{\alpha}{2},$ 可得

推论 5 在 $n$ 维单形 $\Omega_{1},\Omega _{2},\Omega_{3}$ 中, 对 $\alpha>0,$ 有

$\sum\limits_{1\leq i<j\leq n+!}\frac{\rho_{1ij}\rho_{2ij}}{(d_{3i}d_{3j})^{\alpha} }\geq(n!)^{2(1-\alpha)/n}n^{\alpha+1}(n+1)^{[(n+1)\alpha+n-1]/n}\left( \frac{V_{1}V_{2}}{V_{3}^{2\alpha}}\right) ^{1/n}$

(3.24)

$\sum\limits_{1\leq i<j\leq n+1}\frac{\rho_{1ij}\rho_{2ij}}{(h_{3i}h_{3j})^{\alpha} }\geq(n!)^{2(1-\alpha)/n}n^{\alpha+1}(n+1)^{[(1-n)\alpha+n-1]/n}\left( \frac{V_{1}V_{2}}{V_{3}^{2\alpha}}\right) ^{1/n},$

(3.25)

$\sum\limits_{1\leq i<j\leq n+1}\rho_{1ij}\rho_{2ij}(S_{3i}S_{3j})^{\alpha} \nonumber\\ \geq (n!)^{2(1-\alpha)/n}n^{3\alpha+1}(n+1)^{[(1-n)\alpha+n-1]/n}(V_{1} V_{2})^{1/n}(V_{3}^{\alpha})^{2(n-1)/n},$

(3.26)

$\sum\limits_{1\leq i<j\leq n+1}\frac{\rho_{1ij}\rho_{2ij}}{[(R_{3}^{2}-R_{3i} ^{2})(R_{3}^{2}-R_{3j}^{2})]^{\alpha}}\nonumber\\ \geq (n!)^{2(1-2\alpha)/n}n^{2\alpha +1}(n+1)^{[2(n+1)\alpha+n-1]/n}\left( \frac{V_{1}V_{2}}{V_{3}^{4\alpha} }\right) ^{1/n},$

(3.27)

$\sum\limits_{1\leq i<j\leq n+1}\frac{\rho_{1ij}\rho_{2ij}}{(S_{3i}S_{3j})^{\alpha} }\nonumber\\ \geq (n!)^{(1+\alpha)/n}n^{(n-1)\alpha+1}(n+1)^{(n-1)(1-n\alpha)/n}(V_{1} V_{2})^{1/n}\left( \frac{1}{R_{3}^{\alpha}}\right) ^{2(n-1)},$

(3.28)

$\sum\limits_{1\leq i<j\leq n+1}\rho_{1ij}\rho_{2ij}(l_{3i}l_{3j})^{\alpha} \nonumber\\ \geq(n!)^{2(1+\alpha)/n}n^{1-\alpha}(n+1)^{(n-1)(1+\alpha)/n}(n-1)^{-\alpha }(V_{1}V_{2}V_{3}^{2\alpha})^{1/n},$

(3.29)

$\sum\limits_{1\leq i<j\leq n+1}\rho_{1ij}\rho_{2ij}(r_{3i}r_{3j})^{\alpha} \nonumber\\ \geq (n!)^{2/n}n(n+1)^{[n(1+\alpha)-1]/n}(n-1)^{-\alpha}(V_{1}V_{2})^{1/n} r_{3}^{2\alpha},$

(3.30)

$\sum\limits_{1\leq i<j\leq n+1}\rho_{1ij}\rho_{2ij}(r_{3i}^{\prime}r_{3j}^{\prime })^{\alpha}\nonumber\\ \geq(n!)^{2/n}n(n+1)^{(n+2n\alpha-1)/2}(n-1)^{-2\alpha}(V_{1} V_{2})^{1/n}r_{3}^{2\alpha},$

(3.31)

$\sum\limits_{1\leq i<j\leq n+1}\frac{\rho_{1ij}\rho_{2ij}}{[(h_{3i}-d_{3i} )(h_{3j}-d_{3j})]^{\alpha}}\nonumber\\ \geq(n!)^{2(1-\alpha)/n}n^{1-\alpha} (n+1)^{[(n+1)\alpha+n-1]/n}\left( \frac{V_{1}V_{2}}{V_{3}^{2\alpha}}\right) ^{1/n},$

(3.32)

$\sum\limits_{1\leq i<j\leq n+1}\frac{\rho_{1ij}\rho_{2ij}}{[(h_{3i}+d_{3i} )(h_{3j}+d_{3j})]^{\alpha}}\nonumber\\ \geq (n!)^{2(1-\alpha)/n}n^{1+\alpha} (n+1)^{[(n+1)\alpha+n-1]/n}(n+2)^{-2\alpha}\left( \frac{V_{1}V_{2}} {V_{3}^{2\alpha}}\right) ^{1/n},$

(3.33)

当且仅当单形 $\Omega_{1},\Omega_{2},\Omega_{3}$ 均为正则单形且 $P_{3}$ 为 $\ \Omega_{3}$ 的中心时, (3.24), (3.32), (3.33) 式取等号, 当且仅当单形 $\Omega_{1},\Omega_{2},\Omega_{3}$ 均为正则单形时, (3.25)-(3.31) 式取等号.

在定理 1, 2, 3和推论 1, 2, 3中, 取 $\Omega_{2}$ 为 $\Omega_{1},$ $\Omega_{4}$ 为 $\Omega_{3}$ $(\alpha=\beta$ 为 $\dfrac{\alpha}{2}),$ 可 得联系两个单形的一些新的不等式.

注 1 值得指出的是, 由定理的证明不难看出, 只要将 $(S_{i} S_{j})^{\alpha},$ $(l_{i}l_{j})^{\alpha},$ $(r_{i}r_{j})^{\alpha},$ $(r_{i}^{\prime}r_{j}^{\prime})^{\alpha},$ $(d_{i}d_{j})^{-\alpha},$ $(h_{i}h_{j})^{-\alpha},$ $(S_{i}S_{j})^{-\alpha}$, $[(R^{2}-R_{i}^{2} )(R^{2}-R_{j}^{2})]^{-\alpha},$ $[(h_{i}-d_{i})(h_{j}-d_{j})]^{-\alpha},$ $[(h_{i}+d_{i})(h_{j}+d_{j})]^{-\alpha}$ 这十式两两组合, 还有一大批类似定理 1, 2, 3和推论 1~6的联系四个单形或三个单形以及两个单形的一些 新的不等式, 如

$\sum\limits_{1\leq i<j\leq n+1}\frac{\rho_{1ij}\rho_{2ij}(r_{3i}r_{3j})^{\alpha} }{(d_{4i}d_{4j})^{\beta}}\nonumber\\ \geq(n!)^{\frac{2(1-\beta)}{n}}n^{1+\beta}(n+1)^{\frac{n(\alpha +\beta+1)+\beta-1}{n}}(n-1)^{-\alpha}\left( \frac{V_{1}V_{2}}{V_{4}^{2\beta} }\right) ^{1/n}r_{3}^{2\alpha},$

(4.1)

$\sum\limits_{1\leq i<j\leq n+1}\rho_{1ij}\rho_{2ij}(l_{3i}l_{3j})^{\alpha} (r_{4i}^{\prime}r_{4j}^{\prime})^{\beta}\nonumber\\ \geq(n!)^{\frac{2(\alpha-1)}{n}}n^{1-\alpha}(n+1)^{\frac{(n-1)(1-\alpha )+2n\beta}{n}}(n-1)^{-2\beta}(V_{1}V_{2}V_{3}^{2\alpha})^{1/n}r_{4}^{2\beta},$

(4.2)

当且仅当单形 $\Omega_{1},\Omega_{2},\Omega_{3},\Omega_{4}$ 均为正则单形且 $\ \ P_{4}$ 为 $\Omega_{4}$ 的中心时, (4.1) 式取等号, (4.2) 式取等号的条件 同推论 1.

注 2 (3.34)-(3.37) 以及 (3.42), (3.43) 各式是文 [5] 结果的指数推广.

最后, 我们提出如下猜想.

猜想 在$n$ 维单形 $\Omega_{1},\Omega_{2},\Omega _{3},\Omega_{4}$ 中, 对 $\alpha,\beta>0,$ 证明或否定

$\sum\limits_{1\leq i<j\leq n+1}\frac{\rho_{1ij}\rho_{2ij}}{(l_{3i}l_{3j})^{\alpha }(l_{4i}l_{4j})^{\beta}}\nonumber\\ \geq (n!)^{2(1-\alpha-\beta)/n}n^{\alpha+\beta +1}(n+1)^{[(1-n)(\alpha+\beta)+n-1]/n}\left( \frac{V_{1}V_{2}}{V_{3} ^{2\alpha}V_{4}^{2\beta}}\right) ^{1/n}.$

(4.3)

(4.3) 式取等号的条件同推论 1.