I don't work with analysis a great deal, but based on the fact that single-variate polynomials are either constant or have a finite number of local maxima and minima points, I reasoned that this would likely be the case for multivariate polynomials as well. I couldn't prove this to myself, or come up with a counterexample, so I kept silent (Proverbs 17:28). Had I been correct, the puzzle would be answered in the negative, since the global minimum of P would have to be one of a finite number of local minima, so that the image of P would take on a global minimum value at that point, whereas the positive reals has no minimum. But the answer is interesting. I could follow the algebraic reasoning, but I wanted a geometric understanding. So I thought about it, and here is what I came up with: Suppose you have polynomial p(x, y). Then p(x, y)^2 >= 0, so in R3, z = p(x, y)^2 lies above the xy-plane, except where it intersects the xy-plane at p(x, y) = 0. In the case of p(x, y) = x, we see that z = p(x, y)^2 = x^2 lies above the xy-plane, touching it only in the line x = 0. Likewise, for q(x, y) = xy+1, z = q(x, y)^2 = (xy+1)^2 lies above the xy-plane, touching it only on the hyperbola xy+1 = 0. Since the line and hyperbola are disjoint, p(x,y)^2 and q(x,y)^2 are never simultaneously 0, so P(x,y) = p(x,y)^2 + q(x,y)^2 is never 0. Finally, there are points (x,y) on the hyperbola at any desired distance |x| from the line. For these points we have q(x,y) = 0, so P(x,y) = p(x,y)^2 + q(x,y)^2 = p(x,y)^2 = x^2 takes on any desired positive value. On 7/14/2012 3:07 PM, Mike Stay wrote:
P(x, y) = x^2 + (xy + 1)^2
This is clearly nonnegative, and can be made arbitrarily small by letting x -> 0, y = -1/x. Since y can't be infinite, P(x, y) can't be zero.