The question simply asks for the set of values of $\lambda$ so that $(\ln(x^2-2x+2))^2+\lambda \ln(x^2-2x+2)+1\gt 0 \forall x\gt 1$. Since $x^2-2x+2=(x-1)^2+1$. So $\ln(x^2-2x+2)$ is monotonic increasing for $x\in (1,\infty)$. Also $y=\ln(x^2-2x+2)\gt 1$. The inequality gets reduced to $y^2+\lambda y+1\gt 0 \cap y\gt0$. Now it is obvious that the inequality holds for $\lambda \ge 0$. For the other possible values of $\lambda$, set $\lambda^2 -4 \lt 0$. This gives $\lambda \in (-2,2)$. So $\lambda \in (-2,\infty)$.
While this is correct, I want to understand how to prove fundamentally rather than relying on observation that the inequality is true for $\lambda \ge 0$. The discriminant only gives $\lambda \in (-2,2)$. What other condition can be applied to get the remaining set of values $\lambda$ can take? Thanks.
$y^2+\lambda y + 1$ is a convex parable, so if you want that it is positive for $y\ge 0$, then it needs to have no solutions, or all the solutions must be negative.
In the first case, you need a negative discriminant, that is $\lambda^2 < 4$. In the second case, you can solve the quadratic and impose that all the roots are negative, that is $$ -\lambda - \sqrt{\lambda^2-4}\le -\lambda + \sqrt{\lambda^2-4}<0 \implies 0\le\sqrt{\lambda^2-4}<\lambda $$ so $\lambda$ must be at least positive (you could have found the same through Vieta's formulae).