((x²(3-x))/x²-8x+16)≤0 inequation

Given the inequality:
$$\left(- 8 x + \frac{x^{2} \left(3 - x\right)}{x^{2}}\right) + 16 \leq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(- 8 x + \frac{x^{2} \left(3 - x\right)}{x^{2}}\right) + 16 = 0$$
Solve:
$$x_{1} = \frac{19}{9}$$
$$x_{1} = \frac{19}{9}$$
This roots
$$x_{1} = \frac{19}{9}$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} \leq x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + \frac{19}{9}$$
=
$$\frac{181}{90}$$
substitute to the expression
$$\left(- 8 x + \frac{x^{2} \left(3 - x\right)}{x^{2}}\right) + 16 \leq 0$$
$$\left(- \frac{8 \cdot 181}{90} + \frac{\left(\frac{181}{90}\right)^{2} \left(3 - \frac{181}{90}\right)}{\left(\frac{181}{90}\right)^{2}}\right) + 16 \leq 0$$

9/10 <= 0

but

9/10 >= 0

Then
$$x \leq \frac{19}{9}$$
no execute
the solution of our inequality is:
$$x \geq \frac{19}{9}$$

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