Strict inequality versus $\leq$ inequality

Question

Suppose that I know for a real number $x$, the following inequality holds $$ -1\leq x\leq 1\tag{1} $$ Why for $n\in\mathbb{N}$, we have $$ -\frac{1}{n}\leq\frac{x}{n+2}\leq\frac{1}{n}\tag{2} $$ ?

If I divide each sides of Inequality (1) by $n+2$, then I get $$ -\frac{1}{n+2}\leq\frac{x}{n+2}\leq\frac{1}{n+2}\tag{3} $$ I also know that since $n+2>n$, then $$ \frac{1}{n+2}<\frac{1}{n} $$ So I can rewrite Inequality (3) as

$$ -\frac{1}{n}<-\frac{1}{n+2}\leq\frac{x}{n+2}\leq\frac{1}{n+2}<\frac{1}{n}\tag{4} $$ So now Inequality (4) can be written as

$$ -\frac{1}{n}<\frac{x}{n+2}<\frac{1}{n} $$ which is different from Inequality (2)!

Answer 1

$$ -\frac{1}{n}<-\frac{1}{n+2}\leq\frac{x}{n+2}\leq\frac{1}{n+2}<\frac{1}{n}\tag{4} $$

implies

$$ -\frac{1}{n}\leq\frac{x}{n+2}\leq\frac{1}{n} $$

as well.

Though your proof illustrates that equality can't be attained.

It's like there is nothing wrong with writing $2 \le 10$ though we know that $2 < 10$.

Answer 2

If $a< b$ is true, $a\le b$ is also true.

Answer 3

$-1\leq x\leq 1$ is the same as saying that $|x|\leq 1$. Since $0<\frac{n}{n+2}<1$ for $n\in\mathbb{N}$ it follows that $$|x|\frac{n}{n+2}\leq 1\tag{1}$$

$(1)$ is the same as saying that $$-1\leq\frac{xn}{n+2}\leq 1$$ which again, dividing by $n$, is the same as $$-\frac{1}{n}\leq\frac{x}{n+2}\leq\frac{1}{n}$$

Answer 4

The steps that you have taken above are correct and in fact you have proved a stronger statement than was requested.

A final step to finish your answer would be

$$-\frac{1}{n}<\frac{x}{n+2}<\frac{1}{n}\tag{5}$$

Which implies

$$-\frac{1}{n}\le\frac{x}{n+2}\le\frac{1}{n}\tag{6}$$