16. (15 points) Let the function
$$
f(x)=x^{2}-\left(k^{2}-5 a k+3\right) x+7(a, k \in \mathbf{R}) \text {. }
$$

For any $k \in[0,2]$, if $x_{1}, x_{2}$ satisfy
$$
x_{1} \in[k, k+a], x_{2} \in[k+2 a, k+4 a] \text {, }
$$

then $f\left(x_{1}\right) \geqslant f\left(x_{2}\right)$. Find the maximum value of the positive real number $a$.
Three, 16. According to the problem, the axis of symmetry of the function $f(x)$ is
$$
x=\frac{k^{2}-5 a k+3}{2} \text {. }
$$

Therefore, the condition given in the problem is equivalent to, for any $k \in$ $[0,2]$, we have
$$
\begin{array}{l}
\frac{k^{2}-5 a k+3}{2} \geqslant k+\frac{5}{2} a \\
\Leftrightarrow 5 a \leqslant \frac{k^{2}-2 k+3}{k+1} . \\
\text { Also, } \frac{k^{2}-2 k+3}{k+1}=(k+1)+\frac{6}{k+1}-4 \\
\geqslant 2 \sqrt{(k+1) \frac{6}{k+1}}-4=2 \sqrt{6}-4,
\end{array}
$$

with equality holding if and only if $k=\sqrt{6}-1$.
Thus, the maximum value of the positive real number $a$ is $\frac{2 \sqrt{6}-4}{5}$.