$k=\frac{16}{9}$ with equality when $x=y=1/2$ and $z=0$. case 1. if all are negative we have easy calculations to prove it(sum of many positive numbers should be $\geq 0$) case 2. $x\geq0,y\geq0,z\leq0$ is equivalent to positive $x,y,z$ and the inequality becomes $\frac{16}{9}(x^{2}+x+1)(y^{2}+y+1)(z^{2}-z+1)\geq (xyz)^{2}-xyz+1$ set $xy=a^{2}$ constant cauchy gives us $(x^{2}+x+1)(y^{2}+y+1)\geq (a^{2}+a+1)^{2}$ so we need to prove $\frac{16}{9}(a^{2}+a+1)^{2}(z^{2}-z+1)\geq (a^{2}z)^{2}-a^{2}z+1$ if $z\geq 1$ use $z^{2}-z+1=\frac{3}{4}z^{2}+(\frac{z}{2}-1)^{2}\geq \frac{3}{4}z^{2}$, if $z\leq 1$ use $z^{2}-z+1\geq \frac{3}{4}$ plugging in with easy calculations gives the ineq. case 3. $x\geq0,y\geq0,z\leq0$ is again equivalent to positive $x,y,z$ $\frac{16}{9}(x^{2}-x+1)(y^{2}-y+1)\geq \frac{(xyz)^{2}+xyz+1}{(z^{2}+z+1)}$ now if $xy\leq 1$ then $RHS\leq 1$ and it's easy to prove that $\frac{16}{9}(x^{2}-x+1)(y^{2}-y+1)\geq 1$ because $x^{2}-x+1\geq \frac{3}{4}$ if $xy>1$ then $\frac{(xyz)^{2}+xyz+1}{(z^{2}+z+1)}= (xy)^{2}+(1-xy)\frac{1+xy+xyz}{z^{2}+z+1}< (xy)^{2}=\frac{16}{9}(\frac{3}{4}x^{2})(\frac{3}{4}y^{2})\leq$ $\frac{16}{9}(\frac{3}{4}x^{2}+(\frac{x}{2}-1)^{2})(\frac{3}{4}y^{2}+(\frac{y}{2}-1)^{2})=\frac{16}{9}(x^{2}-x+1)(y^{2}-y+1)$ and that should be that... anything shorter??

what a magnificent solution...oh where did you learn to solve inequalities like that? oh you're such a king man...i just love you---

samo da znaš da je editirao poruku koja je glasila: (parafraziram) ti si dokazao da za $k=\frac{16}{9}$ vrijedi ali gdje si pokazao da ne postoji manji $k$... sry for maj bed english, can sambody pliz translate it..

parafraziraš? šta te sve puharićka naučila... sorry guys for this local-language based moments...

I think in case 2 the sign of z should change (in LHS) but not x and y

N.T.TUAN wrote: Find the least real number $ k$ with the following property: if the real numbers $ x, y, z$ are not all positive, then $ k\left(x^{2} - x + 1\right)\left(y^{2} - y + 1\right)\left(z^{2} - z + 1\right)\geq x^2y^2z^2 - xyz + 1$. goc wrote: $ k = \frac {16}{9}$ with equality when $ x = 0$ and $ y=z = \frac{1}{2}$. $ \cdots$anything shorter? WLOG assume $ x\leq0$, then $ \frac{16}{9}\left(x^2+x+1\right)\left(y^2-y+1\right)\left(z^2-z+1\right)-x^2y^2z^2+xyz-1$ $ =\frac{x^2}{63}\left[(7yz-8y-8z+4)^2+42(y-z)^2+6(y+z-4)^2\right]$ $ -\frac{x}{9}\left[4(2yz-y-z+2)^2+3\left(4y^2-5yz+4z^2\right)\right]$ $ +\frac{(2z-1)^2(2y-1)^2}{9}+\frac{(2y-1)^2}{3}+\frac{(2z-1)^2}{3}\geq 0$. Similar problem see http://www.mathlinks.ro/Forum/viewtopic.php?t=140

Define the function $f(t) = t^2-t+1$. Suppose $k\prod (x^2-x+1) \ge (xyz)^2-xyz+1$ for all $x,y,z\in\mathbb{R}$ with $x\le0$. Then for fixed $y,z$, the quadratic \[ P_{y,z}(x) = x^2(kf(y)f(z)-y^2z^2)-x(kf(y)f(z)-yz)+(kf(y)f(z)-1) \] is nonnegative for all $x\le0$. In particular, $P(0) = kf(y)f(z)-1$ and $\lim_{x\to-\infty}\frac{P(x)}{x^2} = kf(y)f(z)-y^2z^2$ must be nonnegative for all $y,z$. Conversely, if $kf(y)f(z)\ge1$ and $kf(y)f(z)\ge y^2z^2$ for all $y,z$, then by AM-GM, $kf(y)f(z)\ge yz$ as well, and we trivially have $P_{y,z}(x) \ge0$ for all $x\le0$ (as the $x^2,-x,1$ coefficients are all nonnegative). Thus the problem reduces to finding the smallest $k$ always satisfying $kf(y)f(z)\ge1$ and $kf(y)f(z)\ge y^2z^2$. As $f(t) = (t-\frac12)^2+\frac34$, the first inequality holds iff $k\ge\frac{16}{9}$. If $yz=0$, the second trivially holds for $k\ge0$; otherwise, since $f(t^{-1}) = f(t)t^{-2}$ for all $t\ne0$, it holds for $k\ge\frac{16}{9}$. We conclude the smallest working value is $\frac{16}{9}$, with equality at $(0,\frac12,\frac12)$ and cyclic shifts. --- We present another proof of $\frac{16}{9} f(x)f(y)f(z) \ge f(xyz)$. It suffices to prove the stronger fact that $\frac43 f(x)f(y) \ge f(xy)$ whenever $x,y$ are not both positive. (WLOG either $x,y,z\le0$, or $xy\le0$; then $\frac{16}{9} f(x)f(y)f(z) \ge \frac43 f(xy)f(z) \ge f(xyz)$.) But $4f(x)f(y)-3f(xy)$ is simply \[ [4f(y)-3y^2]x^2 - [4f(y)-3y]x + [4f(y)-3] = (y-2)^2x^2 - (4y^2-7y+4)x + (2y-1)^2. \]There are many ways to prove this is nonnegative. For instance, WLOG $y\le0$; then it suffices to prove the discriminant \[ (4y^2-7y+4)^2 - 4(y-2)^2(2y-1)^2 = (4y^2-7y+4)^2-(4y^2-10y+4)^2 \]is nonnegative. But $y\le0$ and $8y^2-17y+8 \ge0+0+0$, so we're done.