Ta có
$f(x) \leq 0$
$<-> -(m^2 + 1)x + 2(1 + \sqrt{2}) m + 4 + 2\sqrt{2} \leq 0$
$<-> x\geq \dfrac{2(1 + \sqrt{2})m + 4 + 2\sqrt{2}}{m^2 + 1}$
Để đúng với mọi $x \in [1,2]$ thì
$\dfrac{2(1 + \sqrt{2})m + 4 + 2\sqrt{2}}{m^2 + 1} \geq 2$
$<-> 2(1 + \sqrt{2})m + 4 + 2\sqrt{2} \geq 2m^2 + 2$
$<-> 2m^2 - 2(1 + \sqrt{2})m- 2 - 2\sqrt{2} \leq 0$
$<-> m^2 - (1 + \sqrt{2})m - 1 - \sqrt{2} \leq 0$
$<-> \dfrac{1 + \sqrt{2} - \sqrt{4 + 3\sqrt{2}}}{2} \leq m \leq \dfrac{1 + \sqrt{2} + \sqrt{4 + 3\sqrt{2}}}{2}$