[tex]\dfrac{10}{|x|+1}\leq\dfrac{4}{|x|}\qquad(x\not=0)\\\\10|x|\leq4(|x|+1)\\5|x|\leq2|x|+2\\3|x|\leq2\\|x|\leq\dfrac{2}{3}\\x\leq\dfrac{2}{3} \wedge x\geq-\dfrac{2}{3}\\x\in\left\langle-\dfrac{2}{3},\dfrac{2}{3}\right\rangle\\\\\\x\in\left\langle-\dfrac{2}{3},\dfrac{2}{3}\right\rangle \wedge x\not=0\\\boxed{x\in\left\langle-\dfrac{2}{3},0\right)\cup\left(0,\dfrac{2}{3}\right\rangle }[/tex]
