Помогите решит (x-1)^2<корень из 2*(x-1)

Решите задачу:

$x-1\ \textless \ \sqrt{2(x-1)}\\ x-1 \geq 0\\ x \geq 1\\ (x-1)^2\ \textless \ (\sqrt{2(x-1)})^2\\ x^2-2x+1\ \textless \ 4(x-1)\\ x^2-2x+1\ \textless \ 4x-4\\ x^2-2x+1-4x+4\ \textless \ 0\\ x^2-6x+5\ \textless \ 0\\ x^2-6x+5=0\\ D=36-4*1*5=16\\ x_1= \frac{6+4}{2}=5\\ x_1= \frac{6-4}{2}=1\\ \left \{ {x \geq 1} \atop {(x-5)(x-1)\ \textless \ 0}} \right. \\ x \in (1;5)$