A7 Let a,b,c be positive reals such that abc=1.Prove the inequality ∑2a2+1ab+1a+1≥3
Problem
Source: JBMO 2011 Shortlist A7
Tags: inequalities, Hi
15.05.2016 11:19
2a2+1a=a2+a2+1a≥3a. So we must prove that ∑ab+1a+1≥1.By Cauchy-Schwarz ∑ab+1a+1=∑a2ab+1+a≥(a+b+c)2ab+bc+ca+a+b+c+3.It remains to prove that a2+b2+c2+ab+bc+ca≥a+b+c+3.ab+bc+ca≥33√a2b2c2=3 ⋆a2+b2+c2≥(a+b+c)(a+b+c)3≥3(a+b+c)3=a+b+c ⋆⋆With summing up ⋆ and ⋆⋆ we get desired result. Note: This is also Azerbaijan JBMO TST 2014
15.05.2016 11:32
IstekOlympiadTeam wrote: 2a2+1a=a2+a2+1a≥3a. So we must prove that ∑ab+1a+1≥1.By Cauchy-Schwarz ∑ab+1a+1=∑a2ab+1+a≥(a+b+c)2ab+bc+ca+a+b+c+3.It remains to prove that a2+b2+c2+ab+bc+ca≥a+b+c+3.Which is easy by AM-GM I have another solution but it is Long your solution is the best
15.05.2016 14:08
Murad.Aghazade wrote: A7 Let a,b,c be positive reals such that abc=1.Prove the inequality ∑2a2+1ab+1a+1≥3 See also here http://www.artofproblemsolving.com/community/c6h1077228p4713586
16.02.2023 09:27
Let a=xy, b=yz, c=zx. The desired inequality becomes ∑cyc2x2y2+yxyz+yx+yy≥3which rearranges to ∑cyc(2x2y3+1x)≥3(1x+1y+1z)or ∑cycx2y3≥1x+1y+1zwhich follows by AM-GM since 4⋅x2y3+6⋅y2z3+9⋅z2x3≥1919√x2y3⋅y2z3⋅z2x3=19⋅1x.
01.05.2023 00:59
Setting (a,b,c)=(yz,zx,xy), the inequality simplifies to ∑xy3z2≥∑xy,which is clear say by Rearrangement on {1/z2} and {xy3}.
01.05.2023 18:53
IstekOlympiadTeam wrote: 2a2+1a=a2+a2+1a≥3a. So we must prove that ∑ab+1a+1≥1.By Cauchy-Schwarz ∑ab+1a+1=∑a2ab+1+a≥(a+b+c)2ab+bc+ca+a+b+c+3.It remains to prove that a2+b2+c2+ab+bc+ca≥a+b+c+3.ab+bc+ca≥33√a2b2c2=3 ⋆a2+b2+c2≥(a+b+c)(a+b+c)3≥3(a+b+c)3=a+b+c ⋆⋆With summing up ⋆ and ⋆⋆ we get desired result. Note: This is also Azerbaijan JBMO TST 2014 Nice, last step follows from WAM-WHM.
Attachments:

29.08.2023 17:05
Note that ∑cyc2a2+1ab+1a+1=∑cyc2a2bc+b+1+∑cyc1bc+b+1We now consider the individual parts. Claim: The first part is at least 2. Proof. By Titu's, ∑cyc2a2bc+b+1≥2(a+b+c)2ab+bc+ca+(a+b+c)(abc)13+3(abc)23Substituting in a+b+c=1 and (abc)13≤13, we get that the sum is at least 213+13+13=2.◼ Claim: The second part is exactly 1. Proof. This expands as ∑cyc1ab+a+1=1ab+a+1+11a+b+1+11b+1ab+1=1ab+a+1+aab+a+1+abab+a+1=1◼
21.11.2023 00:09
We first note that a2+b2+c2≥(a+b+c)23≥3√abc⋅(a+b+c)3=a+b+c,2(ab+bc+ca)≥3(abc)2/3+(ab+bc+ca)=3abc+ab+bc+ca. Summing these, we have (a+b+c)2≥3abc+(ab+bc+ca)+(a+b+c). We finish by using Titu's and AM-GM on the numerator: 1≤(a+b+c)2(a+b+c)+(ab+bc+ca)+3abc≤∑ab+bc+abc≤∑13(a2+a2+1a)b+1a+1. ⟹∑2a2+1ab+1a+1≥3. ◼
08.01.2024 06:15
After substituting a=y/x,b=z/y,c=x/z it remains to prove y3x2+z3y2+x3z2≥x+y+z. By AM-GM we have 419(y3x2)+619(z3y2)+919(x3z2)≥(y3x2)419⋅(z3y2)619⋅(x3z2)919=xand summing cyclically finishes.