x²(1-x):x²-4x+4≤0 inequation

Given the inequality:
$$\left(- 4 x + \frac{x^{2} \left(1 - x\right)}{x^{2}}\right) + 4 \leq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(- 4 x + \frac{x^{2} \left(1 - x\right)}{x^{2}}\right) + 4 = 0$$
Solve:
$$x_{1} = 1$$
$$x_{1} = 1$$
This roots
$$x_{1} = 1$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + 1$$
=
$$\frac{9}{10}$$
substitute to the expression
$$\left(- 4 x + \frac{x^{2} \left(1 - x\right)}{x^{2}}\right) + 4 \leq 0$$
$$\left(- \frac{4 \cdot 9}{10} + \frac{\left(\frac{9}{10}\right)^{2} \left(1 - \frac{9}{10}\right)}{\left(\frac{9}{10}\right)^{2}}\right) + 4 \leq 0$$

1/2 <= 0

but

1/2 >= 0

Then
$$x \leq 1$$
no execute
the solution of our inequality is:
$$x \geq 1$$

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