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Inequality challenged (Posted on 2013-03-19) Difficulty: 2 of 5
Given x,y and z are non-negative real numbers satisfying equation:
x + y + z = 1 .
Prove that 0 ≤ yz + zx + xy - 2xyz ≤ 7/27.

Source: Russian IMO .

No Solution Yet Submitted by Ady TZIDON    
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Solution I think I got it. Comment 1 of 1
First off clearly (0,0,1), (0,1,0), and (1,0,0) yield 0,
(1/3,1/3,1/3) yields 7/27.  Not that this proves anything.

y = 1-x-z so we can substitute to give the expression
x(1-x-z) + xz + (1-x-z)z - 2x(1-x-z)z
this can be rewritten as a quadratic in x:
(2z-1)x^2 + (2z^2-3z+1)x + (z-z^2)
(2z-1)x^2 + (2z-1)(z-1)x + (z-z^2)
Which has a min/max at -b/2a = -(z-1)/2
If z=1/2 this expression just evaluates to 1/4 (clearly not a min or max)
If z>1/2 this is always an upward parabola in x (case 1)
If z<1/2 this is always a downward parabola in x (case 2)

Substituting -(z-1)/2 for x gives, with some simplifying
It derivative
-3z^2/2+z/2 is zero when z=0 or z=1/3

So by case 2 this is a maximum value, case 1 does not apply.
Putting 1/3 into the cubic gives 7/27.

  Posted by Jer on 2013-03-20 12:54:06
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