Find the least real number $k$ with the following property: if the real numbers $x$, $y$, and $z$ are not all positive, then \[k(x^{2}-x+1)(y^{2}-y+1)(z^{2}-z+1)\geq (xyz)^{2}-xyz+1.\]
1. Define the function \( f(t) = t^2 - t + 1 \).

2. Suppose \( k \prod (x^2 - x + 1) \ge (xyz)^2 - xyz + 1 \) for all \( x, y, z \in \mathbb{R} \) with \( x \le 0 \). Then for fixed \( y, z \), the quadratic 
   \[
   P_{y,z}(x) = x^2(kf(y)f(z) - y^2z^2) - x(kf(y)f(z) - yz) + (kf(y)f(z) - 1)
   \]
   must be nonnegative for all \( x \le 0 \). 

3. In particular, \( P_{y,z}(0) = kf(y)f(z) - 1 \) and 
   \[
   \lim_{x \to -\infty} \frac{P_{y,z}(x)}{x^2} = kf(y)f(z) - y^2z^2
   \]
   must be nonnegative for all \( y, z \).

4. Conversely, if \( kf(y)f(z) \ge 1 \) and \( kf(y)f(z) \ge y^2z^2 \) for all \( y, z \), then by the Arithmetic Mean-Geometric Mean Inequality (AM-GM), \( kf(y)f(z) \ge yz \) as well. We trivially have \( P_{y,z}(x) \ge 0 \) for all \( x \le 0 \) (as the \( x^2, -x, 1 \) coefficients are all nonnegative).

5. Thus, the problem reduces to finding the smallest \( k \) always satisfying \( kf(y)f(z) \ge 1 \) and \( kf(y)f(z) \ge y^2z^2 \). As \( f(t) = \left(t - \frac{1}{2}\right)^2 + \frac{3}{4} \), the first inequality holds if \( k \ge \frac{16}{9} \). If \( yz = 0 \), the second trivially holds for \( k \ge 0 \); otherwise, since \( f(t^{-1}) = f(t)t^{-2} \) for all \( t \ne 0 \), it holds for \( k \ge \frac{16}{9} \).

6. We conclude the smallest working value is \( \frac{16}{9} \), with equality at \( (0, \frac{1}{2}, \frac{1}{2}) \) and cyclic shifts.

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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 \( \frac{4}{3} f(x)f(y) \ge f(xy) \) whenever \( x, y \) are not both positive. (Without loss of generality, either \( x, y, z \le 0 \), or \( xy \le 0 \); then \( \frac{16}{9} f(x)f(y)f(z) \ge \frac{4}{3} f(xy)f(z) \ge f(xyz) \).)

7. But \( 4f(x)f(y) - 3f(xy) \) is simply
   \[
   [4f(y) - 3y^2]x^2 - [4f(y) - 3y]x + [4f(y) - 3] = (y - 2)^2 x^2 - (4y^2 - 7y + 4)x + (2y - 1)^2.
   \]
   There are many ways to prove this is nonnegative. For instance, without loss of generality, \( y \le 0 \); 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 \( y \le 0 \) and \( 8y^2 - 17y + 8 \ge 0 \), so we're done.

\(\blacksquare\)

The final answer is \( \boxed{ \frac{16}{9} } \)