6. The solution set of the inequality $\sqrt{x^{2}-5 x+4} \leqslant x-5$ is
6. $\{x \mid x \leqslant 1$ or $x \geqslant 4\}$.
$\left\{\begin{array}{l}x^{2}-5 x+4>0 \\ x-5 \geqslant 0 \\ x^{2}-5 x+4>(x-5)^{2}\end{array}\right.$ or $\left\{\begin{array}{l}x^{2}-5 x+4 \geqslant 0 \\ x-5<0,\end{array}\right.$ we get $x \leqslant 1$ or $x \geqslant 4$.