logx2(x+1)2≤1 inequation

Given the inequality:
$$2 \cdot 2 \log{\left(x \right)} \left(x + 1\right) \leq 1$$
To solve this inequality, we must first solve the corresponding equation:
$$2 \cdot 2 \log{\left(x \right)} \left(x + 1\right) = 1$$
Solve:
$$x_{1} = 1.12485600446694$$
$$x_{1} = 1.12485600446694$$
This roots
$$x_{1} = 1.12485600446694$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + 1.12485600446694$$
=
$$1.02485600446694$$
substitute to the expression
$$2 \cdot 2 \log{\left(x \right)} \left(x + 1\right) \leq 1$$
$$2 \cdot 2 \log{\left(1.02485600446694 \right)} \left(1 + 1.02485600446694\right) \leq 1$$

0.198858024543568 <= 1

the solution of our inequality is:
$$x \leq 1.12485600446694$$

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