Lagrange’s Identity and Minkowski’s Conjecture Các giấy tờ chứng minh quy nạp của bất đẳng thức Cauchy sử dụng nhận dạng đa thức (a2 + a2) (b2 + b2) = (a1 b1 + a2 b2) 2 + (b2 a1 a2 b1) 2, 1 2 1 2 () nhưng điều đó chứng minh không có nỗ lực để khai thác công thức này với đầy đủ. Đặc biệt, chúng ta hoàn toàn bỏ qua (b2 a1 a2 b1) nhiệm kỳ 2, ngoại trừ lưu ý rằng nó phải được không âm. Để chắc chắn, bất kỳ sự bất bình đẳng phải. | 3 Lagrange s Identity and Minkowski s Conjecture The inductive proof of Cauchy s inequality used the polynomial identity a2 a2 b2 b2 ab a b 2 ai 2 a2b1 2 but that proof made no attempt to exploit this formula to the fullest. In particular we completely ignored the term a1b2 a2b1 2 except for noting that it must be nonnegative. To be sure any inequality must strike a compromise between precision and simplicity but no one wants to be wasteful. Thus we face a natural question Can one extract any useful information from the castaway term One can hardly doubt that the term a1b2 a2b1 2 captures some information. At a minimum it provides an explicit measure of the difference between the squares of the two sides of Cauchy s inequality so perhaps it can provide a useful way to measure the defect that one incurs with each application of Cauchy s inequality. The basic factorization also tells us that for n 2 one has equality in Cauchy s inequality exactly when a1b2 a2b1 2 0 so assuming that b1 b2 0 0 we see that we have equality if and only if a1 a2 and b1 b2 are proportional in the sense that a1 Ab1 and a2 Ab2 for some real A. This observation has far-reaching consequences and the first challenge problem invites one to prove an analogous characterization of the case of equality for the n-dimensional Cauchy inequality. Problem On Equality in Cauchy s Bound Show that if b1 b2 . . bn 0 then equality holds in Cauchy s inequality if and only if there is a constant A such that ai Abị for all i 1 2 . . n. Also as before if you already know a proof of this fact you are invited to find a new one. 37 38 Lagrange s Identity and Minkowski s Conjecture Passage to a More General iDENTiTy Since the identity provides a quick solution to Problem when n 2 one way to try to solve the problem in general is to look for a suitable extension of the identity to n dimensions. Thus if we introduce the quadratic polynomial Qn Qn a1 a2 . an b1 b2 . bn that is given by the .
