16. Let $f(x)=k\left(x^{2}-x+1\right)-x^{4}(1-x)^{4}$. If for any $x \in[0,1]$, we have $f(x) \geqslant 0$, then the minimum value of $k$ is $\qquad$ .
16. $\frac{1}{192}$.

It is easy to see that $k \geqslant \frac{x^{4}(1-x)^{4}}{x^{2}-x+1}$.
Notice that
$$
x^{2}-x+1=\left(x-\frac{1}{2}\right)^{2}+\frac{3}{4} \geqslant \frac{3}{4} \text {. }
$$

Therefore, when $x=\frac{1}{2}$, $x^{2}-x+1$ reaches its minimum value $\frac{3}{4}$. At this point, $x^{4}(1-x)^{4}$ reaches its maximum value $\left(\frac{1}{2}\right)^{8}$, so the minimum value of $k$ is $\frac{1}{192}$.