Find the smallest $\alpha$ such that, for all $x,y,z$, $\alpha\,\left(x^2-x+1\right)\left(y^2-y+1\right)\left(z^2-z+1\right)\ge(xyz)^2+|xyz|+1$.

Question

Find the smallest $\alpha\in\mathbb{R}$ such that, for all $x,y,z\in\mathbb{R}$, the following inequality holds $$\alpha\,\left(x^2-x+1\right)\left(y^2-y+1\right)\left(z^2-z+1\right)\ge(xyz)^2+|xyz|+1\,.$$

This question is inspired by this thread. According to this related link, it suffices to assume that $x$, $y$, and $z$ are positive. Mathematica indicates that $\alpha=3$ is the lowest possible value, where the sole equality case is $x=y=z=1$. I am looking for a nice solution.

Answer 1

If $x=y=z=1$ we get $\alpha\geq3$.

But for $\alpha=3$ it's enough to prove our inequality for non-negatives $x$, $y$ and $z$.

Since $3(x^2-x+1)^3-(x^6+x^3+1)=(x-1)^4(2x^2-x+2)\geq0$,

our inequality follows from Holder:

$$3\prod_{cyc}(x^2-x+1)\geq\sqrt[3]{\prod\limits_{cyc}(x^6+x^3+1)}\geq x^2y^2z^2+xyz+1$$

Done!

Answer 2

Here is my brute-force attempt. We only need to show that $\alpha=3$ works. It can be easily seen, using Jensen's Inequality and the convexity of $t\mapsto \ln\big(\exp(2t)-\exp(t)+1\big)$ for $t\in\big[\ln(2-\sqrt{3}),\ln(2+\sqrt{3})\big]$, that $$\left(x^2-x+1\right)\left(y^2-y+1\right)\left(z^2-z+1\right)\geq \left(\left(\sqrt[3]{xyz}\right)^2-\sqrt[3]{xyz}+1\right)^3$$ whenever $2-\sqrt{3}\leq x,y,z\leq 2+\sqrt{3}$. Thus, it may be worthwhile to first justify that $$3\,\left(\left(\sqrt[3]{xyz}\right)^2-\sqrt[3]{xyz}+1\right)^3\geq (xyz)^2+xyz+1\,,$$ if $2-\sqrt{3}\leq x,y,z\leq 2+\sqrt{3}$. In other words, we may simply show that $$3\,\left(t^2-t+1\right)^3\geq t^6+t^3+1\tag{1}$$ for all real numbers $t$.

Maybe, the cases where $x$, $y$, or $z$ is not in the interval $\big[2-\sqrt{3},2+\sqrt{3}\big]$ can be handled separately. In fact, Mathematica verifies that, if $0\leq x\leq \frac{1}{3}$ (noting that $\frac{1}{3}>2-\sqrt{3}$), then $$\begin{align} 3\,\left(x^2-x+1\right)\left(y^2-y+1\right)\left(z^2-z+1\right) &\geq \frac{7}{3}\,\left(y^2-y+1\right)\left(z^2-z+1\right) \\&> \left(\frac{yz}{3}\right)^2+\left(\frac{yz}{3}\right)+1 \\&\geq (xyz)^2+xyz+1\,,\end{align}\tag{2}$$ for all $y,z\geq 0$. Similarly, if $x\geq 3$ (noting that $3<2+\sqrt{3}$), then Mathematica also says that $$\begin{align} 3\,\left(x^2-x+1\right)\left(y^2-y+1\right)\left(z^2-z+1\right) &\geq \frac{7}{3}\,x^2\left(y^2-y+1\right)\left(z^2-z+1\right) \\&> x^2\left(\left(yz\right)^2+\left(\frac{yz}{3}\right)+\left(\frac{1}{3}\right)^2\right) \\&\geq (xyz)^2+xyz+1\,,\end{align}\tag{3}$$ for all $y,z\geq 0$. Thus, it may be simpler to tackle (1), (2), and (3) separately.

I have just found out that (1) follows from $$3\left(t^2-t+1\right)^3-\left(t^6+t^3+1\right)=(t-1)^4\left(2t^2-t+2\right)\geq 0$$ for all $t\in\mathbb{R}$. We are now left with the two inequalities $$\frac{7}{3}\,\left(y^2-y+1\right)\left(z^2-z+1\right)> \left(\frac{yz}{3}\right)^2+\left(\frac{yz}{3}\right)+1$$ and $$\frac{7}{3}\,\left(y^2-y+1\right)\left(z^2-z+1\right)> \left(yz\right)^2+\left(\frac{yz}{3}\right)+\left(\frac{1}{3}\right)^2$$ for all $y,z\geq 0$ (in fact, they should be true for all $y,z\in\mathbb{R}$). If these two inequalities hold, then the proof is complete and there is indeed only one equality case—the case $x=y=z=1$.

I shall now prove that $$\frac{7}{3}\,\left(y^2-y+1\right)\left(z^2-z+1\right)\geq \left(yz\right)^2+\left(\frac{yz}{3}\right)+1\,,\tag{*}$$ for which the only equality case is $y=z=1$. The discriminant with respect to $y$ of the polynomial $$7\,\left(y^2-y+1\right)\left(z^2-z+1\right)-3(yz)^2-yz-3=\left(4z^2-7z+7\right)y^2-\left(7z^2-6z+7\right)y+\left(7z^2-7z+6\right)$$ is $$\left(7z^2-6z+7\right)^2-4\left(4z^2-7z+7\right)\left(7z^2-7z+6\right)=-7(z-1)^2\left(9z^2-14z+9\right)\geq 0$$ for all $z\in\mathbb{R}$. Since $4z^2-7z+7>0$ for all $z\in\mathbb{R}$, it is guaranteed that $$7\,\left(y^2-y+1\right)\left(z^2-z+1\right)-3(yz)^2-yz-3\geq 0\,,$$ and the claim follows. That is, $$\frac{7}{3}\,\left(y^2-y+1\right)\left(z^2-z+1\right)\geq \left(yz\right)^2+\left(\frac{yz}{3}\right)+1>\left(yz\right)^2+\left(\frac{yz}{3}\right)+\left(\frac{1}{3}\right)^2$$ and, since the equality case of (*) satisfies $yz\neq 0$, the inequality $$\frac{7}{3}\,\left(y^2-y+1\right)\left(z^2-z+1\right)\geq \left(yz\right)^2+\left(\frac{yz}{3}\right)+1 \geq \left(\frac{yz}{3}\right)^2+\left(\frac{yz}{3}\right)+1$$ holds without an equality case.