Is $3\le 11$ a true statement? $$3(x+1) \le  3x+11$$
simplified: $3 \le 11$.
What is $x$ solution in bracket form?
I think that the expected answer is $(-\infty,\infty)$ a true statement, but it does not feel quite right to me because $3$ is always inferior to $11$ but never equal to it, the sign used in the inequality is $\le $.
What is the right solution in bracket form $(-\infty,\infty)$ or $\emptyset$?

The relation $\le$ stands for less than or equal.
That is an inclusive or, thus if any of the two components less than, or equal is true then the
compound statement is true.
For example, $3\le5$ is true as well as $5\le5$
The solution for $$3(x+1) ≤ 3x+11$$ is therefore $x\in (-\infty, \infty )$
