The given inequality is $|x^2-ax+1|<3(x^2+x+1)$. The question is:
For which values of $a$, every $x$ is a solution?
I am trying to solve it by making the graphics of the two sides of the inequality but have hard time doing so.
2026-03-27 09:37:50.1774604270
Nonstandard inequality with parameter and absolute value
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Hint: Consider the cases $$x^2-ax+1\geq 0$$ and $$x^2-ax+1<0$$ In the first case you have to consider the inequality: $$x^2-ax+1<3(x^2+x+1)$$ and so on.