Find the minimum of a three variate function

Question

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 $m$?

Answer 1

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 $$m=\tfrac{2062+\sqrt[3]{4420439038+12661425\sqrt{120585}}+\sqrt[3]{4420439038-12661425\sqrt{120585}}}{105}=42.04956...$$