Let a,b,c be positive real numbers satisfying a+b+c=3.Prove that ∑syma2(b+c)3≥38
Problem
Source: Greek TST 2012,Pr.3
Tags: inequalities proposed, algebra, Inequality
28.08.2014 17:19
gavrilos wrote: Let a,b,c be positive real numbers satisfying a+b+c=3.Prove that ∑syma2(b+c)3≥38 By C-S ∑cyca2(b+c)3≥(a2+b2+c2)2∑cyca2(b+c)3. Hence, it remains to prove that 8(a2+b2+c2)2(a+b+c)≥9∑cyca2(b+c)3, which is fifth degree. Hence, by uvw we need to check two cases: 1. b=1 and c→0+, which gives something obvious; 2. b=c=1, which gives (a−1)2(4a3+16a2+35a+23)≥0.
28.08.2014 17:35
It is simpler arqady By C-S and Nesbitt we have that 2(a+b+c)∑cyca2(b+c)3≥(∑cycab+c)2≥94
28.08.2014 17:37
Let f(x)=x2(3−x)3. f is convex over [0,3], then by Jensen, we have: f(a)+f(b)+f(c)≥3f(a+b+c3)=38
28.08.2014 17:48
Nice your proofs, silouan and bel.jad5! See also here: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3567758#p3567758
27.11.2021 07:25
Jenson inequality
27.11.2021 07:55
Tangent line also works here. ∑a2(3−a)3≥∑(7a16−516)=38
23.10.2022 03:13
Let a,b,c be positive real numbers. Prove that a2(b+c)3+b2(c+a)3+c2(a+b)3≥98(a+b+c)
06.10.2023 18:43
06.10.2023 19:51
sqing wrote: Let a,b,c be positive real numbers. Prove that a2(b+c)3+b2(c+a)3+c2(a+b)3≥98(a+b+c) My proof is proving your question Mr. Sqing.
07.10.2023 02:16
07.10.2023 02:41
Nice problem..... making the following change (I think it was unnecessary, in the end. ) a=3k,b=3m,c=3n, so k+m+n=1 the problems is equivalent to: ∑sym(3k)233(m+n)3=∑sym≥38 ∑symk2(1−k)3≥98 Let the function f=x2(1−x)3, is convex in the interval [0,1], so by jensen's inequality ∑symk2(1−k)3=f(k)+f(m)+f(n)≥3f(k+m+n3)=3f(13)=98 As desired.
11.10.2023 07:04
Version 1- 4 Variable Let a,b,c be positive reels. Then prove that a2(b+c)3+b2(c+d)3+c2(d+a)3+d2(a+b)3≥2a+b+c+d
11.10.2023 07:05
Generalization 2 Let a,b,c be positive reels (k≥1) such that n≥0 is reel. Then prove that ak(b+c)n+k+bk(c+a)n+k+ck(a+b)n+k≥3n+12n+k(a+b+c)n
11.10.2023 07:13
This is also a generalization of Nesbitt's original inequality when you give k=1,n=0. See Nesbitt- Power Generalization Nesbitt- Power Generalization
11.10.2023 07:14
Generalization 3 Let a,b,c,d ve positive reels (k≥1) such that n≥0 is reel. Then prove that ak(b+c)n+k+bk(c+d)n+k+ck(d+a)n+k+dk(a+b)n+k≥2n+k−2(a+b+c+d)n