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本文将给出关于四面体的两个不等式与其证明。定理一若α_i(i=1,2,……,6)、R、r与α_t′(i=1,2,……6)、R′、r′分别表示四面体ABCD与四面体A′B′C′D′的6条棱长和外接球半径、内切球半径,则成立不等式: 144rr′≤sun from i=1 to 6 α_(?)α_(?)′≤16RR′其中左边等号成立的充分必要条件为:两个四面体均为正四面体;右边等号成立的充分必要条件为:两个四面体对应棱长成比例且每一四面体的三对对棱相等。定理二若m_i、h_i(i=1,2,……,6)、R、r与m_i′、h_i′(i=1,2,……,6),R′、r′分别表示四面体ABCD和四面体A′B′C′D′的四条中线、四条高和外接球半径、内切球半径,则成立不等式:
This article will give two inequalities about tetrahedron and its proof. Theorem one if α_i(i=1,2,...,6), R,r and α_t′(i=1,2,...6), R′, r′ respectively represent tetrahedral ABCD and tetrahedron A′ The length of six edges of B’C’D’ and the radius of the circumscribed sphere and the radius of the inscribed sphere are inequality: 144rr’≤sun from i=1 to 6 α_(?)α_(?)’≤16RR’ The necessary and sufficient condition for the establishment of the number is: two tetrahedra are tetragonal; the necessary and sufficient condition for the right equal sign is: two tetrahedra correspond to the length of the prism and the three pairs of tetrahedrons are equal. Theorem 2 If m_i, h_i (i=1, 2,..., 6), R, r and m_i′, h_i′ (i=1, 2,..., 6), R′, r′ represent tetrahedrons, respectively. ABCD and the four midlines of the tetrahedron A’B’C’D’, the four heights and the radius of the circumscribed sphere, the radius of the inscribed sphere, then establish an inequality: