Solving the inequality $(x^2+3)/x\le 4$ This is the inequality
$$\left(\frac {x^2 + 3}{x}\right) \le 4 $$
This is how I solve it
The $x$ in the left side is canceled and $4x$ is subtracted from both sides.
$$\not{x} \left (\frac {x^2+3} {\not{x}}\right) \le 4x $$
$$ x^2+3 - 4x \le 4x - 4x $$
$$x^2 -4x + 3 \le 0 $$
Then the trinomial is factorized
$$(x-3)(x-1) \le 0 $$
Therefore
$$1\lt x \lt 3 $$
Can someone tell me if I make a mistake, or if the process is good enough. Will you have done something different?

To deal away with fractions, we can multiply both sides of $\left(\frac {x^2 + 3}{x}\right) \le 4$ with $x^2>0$ (here, $x\neq 0$):
$$
(x^2+3)x\leq 4x^2\iff x^3+3x-4x^2\leq 0\iff x(x-1)(x-3)\leq 0.
$$
We want the product of 3 factors above to be nonpositive so there is an odd number of nonnegative factors among $x-3<x-1<x$. This can only happen when 
$$
x-3<x-1<x<0 \quad\text{or}\quad x-3\leq 0\leq x-1<x.
$$
Thus, the answers are $x<0$ or $1\leq x\leq3$.
