【京大1999】(1) の不等式の意味するところは？【不等式の証明】

⭐️ "Oshie Math" allows you to ask Hayashi unlimited questions for a fixed monthly fee. You can try asking questions for free once! https://oshiemath.com/ ⭐️ "Hayashi Math Class," a specialized mathematics school where you can receive direct instruction from Hayashi. Please come for a trial lesson and interview! https://hayashi-math.com/ ✅ Official LINE for students applying to top universities: https://lin.ee/lI7n1SJ Subscriber benefits & live streams for applicants ℹ️ Shunsuke Hayashi's profile https://hayashishunsuke.com/profile/ ・Graduated from Sakae Higashi Junior High School → Chikuma High School → University of Tokyo, Faculty of Science, Department of Physics ・Scored 90% on the second-stage mathematics exam at the University of Tokyo, passing the exam as a current student ・2014 Japan Physics Olympiad Gold Medal ・Placed first place in the 2014 University of Tokyo Physics Mock Exam ℹ️ Please note ・The explanations are Shunsuke Hayashi's own and are not official university information. ・Amazon Associates links will be used when introducing books, etc. This is a proof problem for inequalities from the 1999 Kyoto University entrance exam. The flow is to prove (2) using (1). (1) can be easily proven by eliminating the denominators on both sides (or reducing them to their lowest terms) and considering the difference between the left and right sides. This is a question that all Tokyo University and Kyoto University level candidates will want to answer correctly. Looking at the form of (2), it's clear that (1) would definitely be used, but figuring out how to use it specifically is where the trick is. Looking at (1), we can see that the sum of the fractions is smaller when the a's and b's are the same in the numerator and denominator. This also applies to x1, x2, ..., xn. x1, x2, ..., xn are "rearrangements of 1, 2, ..., n," but the relationship between the subscripts of xk and the value of xk itself is important. Using the inequality in (1), we can determine that the "state where the subscripts and values ​​match," such as x1 = 1, x2 = 2, ..., xn = n, minimizes the sum. Once we can derive this, all we need to do is prove that the sum of 1/(n^2 + 1) is greater than n - 8/5. We can prove this using partial fraction decomposition and evaluation with integrals. This problem had a bit of a Mathematics Olympiad feel rather than a university entrance exam. ▶︎ How to use the inequality in (1) in (2) ▶︎ How to show that the sum of 1/(n^2 + 1) is greater than n - 8/5 The climax is this, and I think it's quite a difficult problem. ---------- 00:00 1999 Kyoto University Science Mathematics [3] 00:44 (1) Subtract the denominators and take the difference 03:26 (1) Summary of the solution 04:25 (2) The meaning of the inequality in (1) 06:53 (2) How to use (1) 13:51 (2) Definition of the pair to be swapped 15:34 (2) Relationship between swapping operations and inequalities 17:38 (2) xk = k after a finite number of operations 19:30 (2) Rephrasing the Tn inequality 22:05 (2) Proof of the Un inequality 26:50 (2) Summary of the solution 30:30 (2) Alternative solution: Evaluating by area 35:04 (2) The limit of Un as n → +∞ 37:35 Conclusion