Find the minimum of a three variate function We consider the function
$$f(x,y,z)=(7(x^2+y^2+z^2)+6(xy+yz+zx))(x^2y^2+y^2z^2+z^2x^2)$$
Find
$$m=\min\{f(x,y,z):xyz=1\}$$
Using the AM-GM inequality it is clear that
$$m_+=\min\{f(x,y,z):xyz=1,x,y,z>0\}=13\times 9=117.$$
But $f(-1,-1,1)=45$ so clearly $m<m_+$. Numerically, it seems that $m\approx 42.0$.
What is the exact value of $m$?
 A: Since $\sum\limits_{cyc}(7x^2+6xy)>0$ and $\sum\limits_{cyc}x^2y^2>0$, we see that the minimal value is non-negative.
Let $m$ be a minimal value.
Thus, $$\sum_{cyc}(7x^2+6xy)\sum_{cyc}x^2y^2\geq mx^2y^2z^2.$$
Let $x+y+z=3u$, $xy+xz+yz=3v^2$, where $v^2$ can be negative and $xyz=w^3$.
Hence, $$(7(9u^2-6v^2)+18v^2)(9v^4-6uw^3)-mw^6\geq0$$ or $g(w^3)\geq0,$ where $$g(w^3)=(7(9u^2-6v^2)+18v^2)(9v^4-6uw^3)-mw^6.$$ 
We see that $g$ is a concave function.
But the concave function gets a minimal value for an extreme value of $w^3$, which happens for an equality case of two variables.
Since $g(w^3)$ is a homogeneous, it's enough to assume $y=z=1$, which gives
$$m=\min_{x\in\mathbb R}\frac{(7x^2+12x+20)(2x^2+1)}{x^2}.$$
We obtain  $$\left(\frac{(7x^2+12x+20)(2x^2+1)}{x^2}\right)'=\frac{4(x-1)(7x^3+13x^2+13x+10)}{x^3},$$
which gives that the minimum occurs when $x$ equal to the real root of the equation $$7x^3+13x^2+13x+10=0,$$
which we can get by the Cardano's formula and we can get an exact  minimal value. 
I got
$$m=\tfrac{2062+\sqrt[3]{4420439038+12661425\sqrt{120585}}+\sqrt[3]{4420439038-12661425\sqrt{120585}}}{105}=42.04956...$$
