Could You check whether this expression is nonnegative? I'm trying to determine if $$\bigl(x+y)^4(y+z)^4(z+x)^4 \geq$$
$$8x^2y^2z^2\bigl((x+y)^2 + (y+z)^2\bigr)\bigl((y+z)^2 + (z+x) ^2\bigr)\bigl((z+x)^2 + (x+y)^2\bigr)$$
for $x,y,z>0$.

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!
