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xyz inequality (Posted on 2015-07-24) Difficulty: 2 of 5

  
If x, y, and z are nonnegative real numbers,
then prove that

   x2 + xy2 + xyz2 ≥ 4xyz - 4
  

See The Solution Submitted by Bractals    
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solution Comment 1 of 1
let
f(x,y,z)=x^2+xy^2+xyz^2-4xyz
we wish to prove f(x,y,z)>=-4

df/dx = 2x+y^2+yz^2-4yz = 0  (1)
df/dy = 2xy+xz^2-4xz = 0 (2)
df/dz = 2xyz-4xy = 0 (3)

from (3) we have
xyz=2xy
first assume x=0, but then f(0,y,z)=0>=-4
thus assume x>0 then we have
yz=2y
assume y=0 then we have
2x=0 which contradicts x>0
thus y>0
which gives us z=2
now we have
2x+y^2+4y-8y=0 (4)
2xy+4x-8x=0 (5)
from (5)
2xy=4x since x>0
y=2
and from (4) we get
2x+4+8-16=0
x=2

f(2,2,2)=-4
thus f(x,y,z)>=-4 for x,y,z>=0

  Posted by Daniel on 2015-07-24 08:45:54
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