Let x,y,z be positive real numbers satisfying x+y+z=xyz. Find the minimum value of
x+y y+z z+x
(------- + ------- + -------)2
1-xy 1-yz 1-zx
Multiply the first term by z/z to get
z(x+y) / (z-xyz)
z(x+y) / (z-(x+y+z))
-z(x+y) / (x+y)
= -z
Similarly for the other two terms.
(-z-x-y)^2
So the expression we are to minimize simplifies to
(x+y+z)^2 or (xyz)^2
The problem becomes:
Minimize (x+y+z); or Minimize (xyz)
while x+y+z=xyz, and
none of xy, xz, or yz can equal 1, and
x,y,z all positive.
Try a short cut:
What if x=y=z? Then x^3 = 3x; x^3 - 3x = 0
x(x+√3)(x-√3) = 0 and of 0, √3, or -√3 solves.
Rejecting zero and negative leaves the only solution (if x=y=z) as
x=y=z = √3; then the sum and product = 3√3
So this is the minimum if x=y=z, but this is not yet a definitive proof.
Expand (x+y+z)^3
= (x^3+y^3+z^3) + 3(x(xy+xz) + y(xy+yz) + z(xz+yz)) + 6xyz
I suspect some algebraic manipulation will lead to the answer, but I haven't had enough coffee yet for that.
Try AM >= GM for the set (x,y,z)
AM: (x+y+z)/3
GM: (xyz)^(1/3)
(x+y+z)/3 >= (xyz)^(1/3)
(x+y+z)^3 >= 27(xyz) but x+y+z=xyz
(xyz)^3 >= 27(xyz)
(xyz)^3 - 27(xyz) >= 0
(xyz)*(xyz + 3√3)*(xyz - 3√3) >= 0
But xyz is always positive; the first 2 terms are positive so the third term must be positive. Therefore (xyz - 3√3) >= 0
xyz >= 3√3
We have one solution where xyz = 3√3 and we have proved that xyz >= 3√3.
Thus xyz = x+y+z = 3√3 is the minimum value.
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Posted by Larry
on 2024-12-28 10:13:53 |
Solution