Solve $\log_{1/3}(x^2-3x+3)≥0$ I want to solve $$\log_{1/3}(x^2-3x+3)≥0$$
Now I know the result is: $x ∈ <1;2>$, but i am not sure how to get it.
My thoughts: $\frac{1}{3}$ to the power of positive number $= (x^2-3x+3)$, now I would solve $x^2-3x+3$ with the help of discriminant to get the points where $x$ is zero, and say the answer is for the positive intervals. Is this correct / is there easier way to do it?

Raise $\frac{1}{3}$ to the power of both sides, and flip the inequality because $\frac{1}{3}^x$ is a decreasing function:
$$
\begin{aligned}
\log_{1/3}(x^2 - 3x + 3 ) &≥ 0\\
x^2 - 3x + 3 &\le \left(\frac{1}{3}\right)^0\\
x^2 - 3x + 3 &\le 1\\
x^2 - 3x + 2 &\le 0\\
\end{aligned}
$$
Now factor to get $(x-1)(x-2) \le 0$. We need a negative and a positive sign to make a negative, hence
$$1 \le x \le 2$$
