Help with this inequality : $-1 \le \frac{1+x^2}{2x} \le 1$ I  have to solve 
$$
-1 \le \frac{1+x^2}{2x} \le 1
$$ 
My attempt at solution:
$$ -1 ≤ (1+x²)/2x ≤ 1 $$
$$ - 1 ≤ (1+x²)/2x  \quad\text{and}\quad  (1+x²)/2x ≤ 1 $$
$$ 0≤(1+x²)/2x + 1   \quad\text{and}\quad (1+x²)/2x - 1   ≤0 $$
i.e. $(x²+2x+1)/2x   \quad\text{and}\quad (x²-2x+1)/2x $
$$ 0≤ (x+1)²/2x \quad\text{and}\quad  (x-1)²/2x  ≤0 $$
since the numerator is positive, the sign depends on the denominator
i.e. $\;0≤x$ and  $x≤0$
$$ 
therefore the answer is $0$ which is wrong as $1$ and $-1$ also satisfy the inequation and zero will make $(1+x²)/2x$ not defined, so where did I go wrong?

One way to solve it is to consider two separate cases (x < 0 and x > 0). Note that x cannot be 0. 
Case 1. When x > 0, we have 
$ – 2x \le 1 + x^2 \le 2x$
I.e. ($0 \le (1 + x)^2$) and ($(1 – x)^2 \le 0$)
I.e. (All x) and (x = 1)
i.e. x = 1
Combining x = 1 with the initial x > 0, we have x = 1.
Similarly, we get x = –1 from case 2.
