Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have posted a related question before (link), but I didn't really get a completely satisfactory answer, and also I believe that I was able to simplify the problem a little. Therefore, I hope that it is appropriate to post again.

Basically, I want to show that

$f(y,z) := 81 y^4+\left(450 y^2+228 y^3+476 y^4\right) z+\left(405+108 y+1692 y^2+504 y^3+884 y^4\right) z^2+\left(1404-168 y+1828 y^2-144 y^3+528 y^4\right) z^3+\left(1872-624 y+672 y^2-576 y^3\right) z^4+\left(1232-192 y+320 y^2\right) z^5+320 z^6$

is positive for any $y > 0, z>0$.

By inspection of the $y$ polynomials, it is easy to see that only the polynomial before $z^4$, $g(y) := 1872-624y + 672y^2-576y^3$, might cause troubles for $y$ sufficiently large. However, even when ignoring the positive terms in $g(y)$, and taking $h(y) := -624y - 576y^3$ instead, numerically it still holds that $f(y,z)$ is strictly positive.

It hope that there is some easy way to show positivity of $f(y,z)$. Any idea would be greatly appreciated. Thanks!

share|cite|improve this question

Maple manages to find all critical points (all digits given are correctly rounded, and it is guranteed that there are no additional solutions). These turn out to be:

  1. y = -1.200130607, z = -1.801192473
  2. y = -1.500000000, z = -1.500000000
  3. y = -1.394584620, z = -1.299681645
  4. y = -1., z = -1.
  5. y = -1.116359746, z = -.8642530441
  6. y = -1.500000000, z = -.7500000000
  7. y = -1.173750738, z = -.5083262416
  8. y = -.7095004638, z = -.2328818417
  9. y = 0., z = 0.

None of these are in the first quadrant. It's straight-forward to check that $f \ge 0$ on the boundary of the the first quadrant, and that $f\to\infty$ as $y^2+z^2 \to \infty$.

Hence, solving a min/max-problem on a large (closed) quarter-disc $K$ in the first quadrant, the extrema must be on the boundary of $K$ and here we know that $f \ge 0$.

The global minimum of $f$ on $K$ must then be $0$, and this is attained only at $(0,0)$, so your $f$ is indeed positive on the open first quadrant

share|cite|improve this answer
Thanks, I appreciate the effort! The problem is that it's numerically clear anyway that $f(y,z)$ must be positive. So finding its critical points numerically doesn't really solve the problem (although the rest of your argument is of course correct). – Martin Feb 8 '13 at 10:36
@Martin The point is that the computation is just "semi-numeric". I didn't use the regular "fsolve" but a root-isolating algorithm that internally uses some variant of interval arithmetic or other branch-and-bound technique that proves correctness of the numerical approxmations. The computation above is a guarantee (with mathematical certainty) that there are no critical points in the first quadrant. – mrf Feb 8 '13 at 11:24

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.