Expert: Abe Mantell - 1/1/2010

Question
QUESTION: What is the absolute minima of this
x^2/(x-1)^2 +y^2/(y-1)^2 + (1/xy)^2 / (1- (1/xy) )^2?

ANSWER: It appears to be 1.


---------- FOLLOW-UP ----------

QUESTION: umm sorry I actually knew that its 1 I need the steps...
I don't understand how to do this D this can you explain it?

Answer
Letting z(x,y)=x^2/(x-1)^2+y^2/(y-1)^2+1/(x^2*y^2*(1-1/(x*y))^2)
the partial derivatives are:
zx=2*x/(x-1)^2-2*x^2/(x-1)^3-2/(x^3*y^2*(1-1/(x*y))^2)-2/(x^4*y^3*(1-1/(x*y))^3)
zy=2*y/(y-1)^2-2*y^2/(y-1)^3-2/(x^2*y^3*(1-1/(x*y))^2)-2/(x^3*y^4*(1-1/(x*y))^3)

Solving zx=0 & zy=0, gives:
{x = 1/4, y = -2} and {x = (1/2)*(3*y-1+sqrt(9*y^2-6*y+1-4*y^3))/y^2, y = y} as the only
real solutions.  Substituting these back into z(x,y) gives z=1 for both cases.
Since it is clear that there is no maximum (z approaches infinity for x->0, y->0, etc.),
with z being bounded below (since all the terms are squared, z cannot be negative) z=1
must be a minimum.

Abe