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I'm trying to determine if $$\bigl(x+y)^4(y+z)^4(z+x)^4 - 8x^2y^2z^2\bigl((x+y)^2 + (y+z)^2\bigr)\bigl((y+z)^2 + (z+x) ^2\bigr)\bigl((z+x)^2 + (x+y)^2\bigr) \ge0 $$

for $x,y,z>0$.

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I think Wolfram Alpha would help you much faster (if you ask nicely)! – Fabian Aug 6 '12 at 16:18
The first part of your text read {(x+y)(y+z)(z+x)}^4: do you mean that all three terms are raised to the 4th power? – enzotib Aug 6 '12 at 16:25
Fabian / Wolfram Alpha do times out. Enzotib / Yes $ (x+y)^4 (y+z)^4 (z+x)^4 $ – guldam Aug 6 '12 at 16:57
Umm.. Would you let me know if you know a nice way to determine? Sine the homogeneousity of the expression, we can just set z=1 and then Mathematica show the graph is above the horizontal plane for x,y>0 but.. – guldam Aug 6 '12 at 16:59
I edited my expression to be more clear. – guldam Aug 6 '12 at 17:15

One way to attack it is to compute a sum-of-squares solution numerically through semidefinite programming, i.e., finding a positive semidefinite matrix $Q$ such that your polynomial $p(x,y,z)$ can be written as $v^TQv$ where $v$ is a suitably selected vector of monomials in $x,y,z$.

The following piece of code computes a sum-of-squares solution in MATLAB by using the toolbox YALMIP. It assumes that you have an efficient semidefinite solver installed, such as SeDuMi or SDPT3

sdpvar s t u
x = s^2;
y = t^2;
z = u^2;
p = (x+y)^4*(y+z)^4*(z+x)^4-8*x^2*y^2*z^2*((x+y)^2+(y+z)^2)*((y+z)^2+(x+z)^2)*((z+x)^2+(x+y)^2);
[diags,v,Q] = solvesos(sos(p));

The solution $Q$ is a 61x61 matrix, close to singular, positive semidefinite matrix (with some block-structure arising due to symmetry exploitation performed in the decomposition). Note though, it is only a numerical indication of positivity. The result is only correct up to roughly 7 or 8 digits, since there is a discrepancy between p and the computed decomposition.


However, although not a true certificate, it might give some hints towards a symbolic decomposition, and hints at what the answer is.

By attacking @Camerons polynomials instead, numerics are more to our advantage

sdpvar s t
a = 2 + s^2;
b = 2 + t^2;
p = ((a+b-2)*a*b)^4-8*(a-1)*(b-1)^2*((a+b-2)^2+a^2)*((a+b-2)^2+b^2)*(a^2+b^2)
[diagnostics,v,Q] = solvesos(sos(p))

This decomposition has a 68x68 matrix Q with smallest eigenvalue ~0.45, and by using some trivial results relating the difference between $p$ and $v^TQv$ with the smallest eigenvalue of $Q$, it can be shown that this decomposition actually proves positivity, despite being approximate(Theorem 4). Alas, everything is still only correct to machine precision etc, i.e., still not a truly valid certificate.

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Updated Version: Call the given polynomial $f(x,y,z)$. Note that this is a symmetric polynomial--that is, if we permute $x,y,z$ in the polynomial, we end up with the same polynomial--so without loss of generality, we may assume that $x,y\geq z$.

Another nice observation is that every term of the expanded polynomial will have the same total degree--namely $12$--so the polynomial is homogeneous, too. This suggests that we might want to rewrite $x=\alpha z$, $y=\beta z$, where $\alpha,\beta\geq 1$ (since $x,y\geq z>0$), for then each term will have a factor of $z^{12}$, and our $3$-variable problem reduces to the (less daunting) $2$-variable problem of determining whether $$\frac{f(\alpha z,\beta z,z)}{z^{12}}\geq 0$$ for all $\alpha,\beta\geq 1$.

To make our lives slightly easier (so we aren't working with so much at once and everything fits on one line), let's set $$p(x,y,z):=\bigl((x+y)(x+z)(y+z)\bigr)^4$$ and $$q(x,y,z):=8x^2y^2z^2\bigl((x+y)^2+(x+z)^2\bigr)\bigl((x+y)^2+(y+z)^2\bigr)\bigl((x+z)^2+(y+z)^2\bigr),$$ so that $f=p-q$. Now, $$p(\alpha z,\beta z,z)=\bigl((\alpha+\beta)(\alpha+1)(\beta+1)\bigr)^4z^{12}$$ and $$q(\alpha z,\beta z,z)=8\alpha^2\beta^2\bigl((\alpha+\beta)^2+(\alpha+1)^2\bigr)\bigl((\alpha+\beta)^2+(\beta+1)^2\bigr)\bigl((\alpha+1)^2+(\beta+1)^2\bigr)z^{12}.$$

These are still messy, so let's perform another substitution for increased simplicity. Set $a:=\alpha+1$, $b:=\beta+1$, so $$p(\alpha z,\beta z,z)=\bigl((a+b-2)ab\bigr)^4z^{12}$$ and $$q(\alpha z,\beta z,z)=8(a-1)^2(b-1)^2\bigl((a+b-2)^2+a^2\bigr)\bigl((a+b-2)^2+b^2\bigr)(a^2+b^2)z^{12}.$$ Hence, we need to show that $$\bigl((a+b-2)ab\bigr)^4-8(a-1)^2(b-1)^2\bigl((a+b-2)^2+a^2\bigr)\bigl((a+b-2)^2+b^2\bigr)(a^2+b^2)\geq 0$$ for all $a,b\geq 2$.

At this point, though, I'm stuck. Perhaps this will give you (or someone else) an idea on how to get the rest of the way--or perhaps Mathematica will give you what you need. Sorry!

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I edited just before to determine not $$>0$$ but $$\ge0$$ .. Sorry for my first mistake. – guldam Aug 6 '12 at 17:17

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