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?

This problem is equivalent to $$\left|\frac{1+x^2}{2x}\right|\le 1,$$ which gives $$(1+x^2)^2\le(2x)^2,$$ which gives $$(1-x)^2(1+x)^2\le 0.$$ Since LHS of last inequality can never be negative, it follows that the only solution will occur when LHS vanishes. Can you continue now?
