A Note On The Neuberg-Pedoe Type Inequalities

In this thesis we have mainly studied the simplex and discussed the Neuberg-Pedoe type inequalities.Let a, b, c and a' V, c' are the edges of â–³ABC and â–³A'B'C', A and â–³' are their areas. The famous Neuberg-Pedoe inequality isa'²{b² + c² - a²) + b'²(c² + a² - b²) + c'²(a² +b² - c²) ≥ 16△△', (1)and the equality holds if and only if AABC is similar to AA'B'C.([2]) Chia kuei Peng got a result as follows: a²(b'² + c'²-a'²)+b²(c'² + a'²-b'²)+c²(a'² + b'²-c'²)≥8(λ'/λ â–³² + λ'/λ â–³'²), (2)and the equality holds if and only if AABC is similar to AA'B'C', A = a² + b² +c², λ' = a'²+b'²+c'²([9])In 1981, Lu Yang and Jingzhong Yang[3] generalized (1) in Euclidean space Eⁿ. And Huaming Su[4], Gangsong Leng[6], Hanfang Zhang[7] generalized (1) and (2). In this thesis we generalize (1) and (2) in Euclidean space Eⁿ based on their works.In Chapter 3 we get two results.Let {A₁,… , A_n+1} are the vertexes of the simplex E, V is the volume of E, and V_{i,i1,…ik} is the volume of the sub-simplex {A_i, A_1i,…A)_ik}, 1 ≤ i, i₁, i₂,…, i_k ≤n + 1 are different, V_{i,i1,…ik} =0 if i= i_j(1 j ≤ κ). LetTheorem 1 : For the simplices S and S' we can get:1 ≤ κ ≤ n +1, and the equality holds if and only if both E and S' are the regular simplices.There are numbers of k dimensional sub-simplices in E. The volume of i-th sub-simplex in S is denoted by V_k,i, 1 ≤ i, ≤ m.Theorem 2 : For the simplices Σ and Σ' we can get:and the equality holds if and only if both E and E' are the regular simplices, where(k + )\ * / nl2 {n, k) = m (m - (n + 1 - A;)) ( -^ n!12(5)In Chapter 4, we discuss two inequalities proved by G. Leng and H. Zhang.Let V be the volume of E = {Aq, A\, â–  ? â–  , An}, V* is the volume of {.Ao, A\, â–  Ai+i, â–  ? ? , A^,}, 0 < i < n. LetF(y Q\ V^ T/28 ( Y^ y29 9T/25A / 2 -t \ ni=o j=o LE is the simplex and 6 > 0.When 0 6 (0, |] G. Leng proved that F(E, 0) > 0 and H. Zhang proved that F(E, 6) > 0 for 0 = 1. The problem is whether F(E, 0) > 0 still holds when 0 G (|, 1). In Chapter 4 we discuss this problem in detail and construct a counterexample to show F(E, 0) > 0 is not correct when 0 G (|, 1).