For arbitrary positive reals a≥b≥c prove the inequality: a2+b2a+b+a2+c2a+c+c2+b2c+b≥(a+b+c)+(a−c)2a+b+c(Anton Trygub)
Problem
Source: 2021 Ukraine NMO 10.3
Tags: algebra, inequalities
02.04.2021 22:23
parmenides51 wrote: For arbitrary positive integers a≥b≥c prove the inequality: a2+b2a+b+a2+c2a+c+c2+b2c+b≥(a+b+c)+(a−c)2a+b+c(Anton Trigub) It's true for any positives a, b and c. C-S of course.
02.04.2021 22:29
bad translation by google, reals and not integers is the correct
02.04.2021 23:28
Solution:
03.04.2021 03:19
Let a,b,c be positive reals. Prove that a2+b2a+b+a2+c2a+c+c2+b2c+b≥a+b+c+(a−c)2a+b+cSolution: ∑cyca2+b2a+b=∑cyca+12∑cyc(a−b)2a+b≥∑cyca+12((a−c)2a+2b+c+(c−a)2c+a)≥∑cyca+(a−c)2a+b+c
07.05.2021 11:59
parmenides51 wrote: For arbitrary positive reals a≥b≥c prove the inequality: a2+b2a+b+a2+c2a+c+c2+b2c+b≥(a+b+c)+(a−c)2a+b+c(Anton Trygub) ∑a2+b2a+b−∑a+b2=(a−b)22(a+b)+(b−c)22(b+c)+(a−c)22(c+a)≥(2a−2c)24(a+b+c)=(a−c)2a+b+c◼
05.08.2022 04:06
For arbitrary positive reals a,b,c prove the inequality: a3+b3a+b+a3+c3a+c+c3+b3c+b≥a2+b2+c2+6(a−c)47(a2+b2+c2)√a3+b3a+b+√a3+c3a+c+√c3+b3c+b≥a+b+c+(a−c)2a+b+chttps://artofproblemsolving.com/community/c6h1791421p11852916
Attachments:
