tailieunhanh - THE CAUCHY – SCHWARZ MASTER CLASS - PART 15

Bắt đầu với giải pháp Cauchy cho Tập thể dục . Bất bình đẳng đầu tiên sau bằng cách áp dụng bất đẳng thức Cauchy {ak} và {bk} một trong những nơi có bk = 1 cho tất cả các k. Cô lập, "1-trick" gần như là tầm thường, nhưng nó là khá tổng quát: tổng hợp tất cả có thể được ước tính theo cách này. Nghệ thuật chứ không phải là một trong các dự đoán khi các ước tính kết quả có thể chứng minh là hữu ích. Đối với vấn đề thứ hai, chúng. | Solutions to the Exercises Chapter 1 Starting with CAucHy Solution for Exercise . The first inequality follows by applying Cauchy s inequality to ak and bk where one takes bk 1 for all k. In isolation this 1-trick is almost trivial but it is remarkably general every sum can be estimated in this way. The art is rather one of anticipating when the resulting estimate might prove to be helpful. For the second problem we apply Cauchy s inequality to the product of ak ano. ak . -L His is a sini zie instance or Hie s ziivving L-Licn. wneie one estimates the sum of the ak by Cauchy s inequality after writing ak as a product ak bk ck. Almost every chapter will make some use of the splitting trick and some of these applications are remarkably subtle. Solution for Exercise . This is another case for the splitting trick one just applies Cauchy s inequality to the sum n . . 1 1 Ì 1 1 1 1 pj aj pj bj Solution for Exercise . The first inequality just requires two applications of Cauchy s inequality according to the grouping ak bkck but one might wander around a bit before hitting on the proof of second inequality. One key to the proof of the second bound comes from noting that when we substitute ak bk ck 1 we get the lackluster bound n2 n3. This suggests the inequality is not particularly strong and it encourages us to look for a cheap shot. One might then think to deal 226 Solutions to the Exercises 227 with the ck factors by introducing ck ck cl c2 ----- cn so the target inequality would follow if we could show . . 1 J 2 y ỳ ak bkck I a J y ỳ bk but this bound is an immediate consequence of the usual Cauchy inequality and the trivial observation that Cfc 1. Solution for Exercise . For part a we note by Cauchy s inequality and the 1-trick that we have S2 12 12 12 f y x z y z _ t 6. x y z x y z x y z For part b we apply Cauchy s inequality to the splitting . x y I z ------------- x y z x yy z x yx z x yx y. Solution for Exercise . From Cauchy s inequality the .