Prove that for all a,b,c positive real numbers a4+1b3+b2+b+b4+1c3+c2+c+c4+1a3+a2+a≥2
Problem
Source: Turkey JBMO TST 2023 Day 2 P1
Tags: inequalities
30.04.2023 14:33
Just note that a4+1≥23⋅(a3+a2+a) (it is equivalent to (a−1)2(3a2+4a+3)≥0), and use AM-GM.
30.04.2023 19:32
My solution from the exam. LHS≥33√(a4+1)(b4+1)(c4+1)(a3+a2+a)(b3+b2+b)(c3+c2+c) Let's prove that a4+1a3+a2+a≥23 3a4+3≥2a3+2a2+2a(a2−1)2+2(a4−a3−a+1)≥0(a2−1)2+2(a−1)2(a2+a+1)≥0The last inequality is true.
01.05.2023 20:26
Using x4+1≥x3+x (equivalent to (x−1)2(x2+x+1)≥0), we get that LHS≥33√∏a3+aa3+a2+a=33√∏(1−a2a3+a2+a).Next, by AM-GM, a3+a≥2a2, so a2/(a3+a2+a)≤1/3. This immediately yields the desired result.
05.05.2023 02:13
egxa wrote: Prove that for all a,b,c positive real numbers a4+1b3+b2+b+b4+1c3+c2+c+c4+1a3+a2+a≥2
Attachments:



28.02.2024 10:01
Orestis_Lignos wrote: Just note that a4+1≥23⋅(a3+a2+a) (it is equivalent to (a−1)2(3a2+4a+3)≥0), and use AM-GM. How did you get, that a4+1≥23⋅(a3+a2+a) (it is equivalent to (a−1)2(3a2+4a+3)≥0)
01.01.2025 12:28
Just open the brackets @above