Prove that for all x,y,z>0, the inequality x+y+z3+31x+1y+1z≥53√xyz16holds. Determine if equality can hold and if so, in which cases it occurs.
Problem
Source: Germany 2015, Problem 6
Tags: inequalities, inequalities proposed, mean, Harmonic Mean
05.05.2017 20:08
Easy uvw.
05.05.2017 20:22
arqady wrote: Easy uvw. i don't think that official solution is with uvw i am pretty sure that there is a simple nice proof
05.05.2017 22:53
The hard part is to figure out and equality case, which is (1,4,4). Then if we denote x=a3,y=4b3,z=4c3 a full expansion and a smooth combination of AM-GM for 90 terms will end up the problem.(I hope I'm right)
05.05.2017 22:57
GGPiku wrote: The hard part is to figure out and equality case, which is (1,4,4). Then if we denote x=a3,y=4b3,z=4c3 a full expansion and a smooth combination of AM-GM for 90 terms will end up the problem.(I hope I'm right) This will never ever work and the reason is simple: There are two more equality cases: (4,1,4) and (4,4,1). So you won't find an AM-GM estimate in which x plays a special role. In fact, one of the official solutions argues indeed essentially with uvw. My favourite solution (which I found during that competition) is a mixing-variables approach: W.l.o.g. let x≥y≥z. Let LHS−RHS=f(x,y,z). Then it is not hard to show that f(x,y,z)≥f(√xy,√xy,z) and from here by homogenity the problem reduces to a one-variable inequality which is then easily established.
19.08.2019 11:58
As Tintarn said x≥y≥z. f(x,y,z)≥f(√xy,√xy,z)⟺(√x−√y)2[(2x+2y−9√xy)z2+xy√xy+z⋅√xy(√x+√y)2]≥0 We have to prove (2x+2y−9√xy)z2+xy√xy+z⋅√xy(√x+√y)2≥0If 2x+2y−9√xy>0 then obviously holds. If 2x+2y−9√xy≤0 it's sufficient to prove for z=0,z=y. First case is clear and second is: 2y2(x−√xy)+y√y(√x−√y)(x+√xy−y)≥0and it's true (x≥y).
19.08.2019 14:06
Tintarn wrote: Prove that for all x,y,z>0, the inequality x+y+z3+31x+1y+1z≥53√xyz16holds. Determine if equality can hold and if so, in which cases it occurs. WLOG, we can assume xyz=2. We need to prove: x+y+z3+31x+1y+1z≥52The inequality is linear on u. According to uvw theorem, we only need to check y=x and z=2x2. In this case, we get: (x−2)2(4x4+x3−12x2+4x+4)6x2(x3+4)≥0which is obviously true. Equality at (x,y,z)=(2,2,12) and their permutations.