Rearranging the inequality we get $\frac{(4(x+2)^2-x^2)^2}{16x^3} \geq 0$ Clearly as the numerator is a perfect square it is nonnegative Do we get that $16x^3 \geq 0$ But $x \neq 0$ as it is in the denominator So the inequality holds for all $x>0$

$x=-4$ or $x=-4/3$ is also possible...

parmenides51 wrote: Solve in real numbers $\frac{(x+2)^4}{x^3}-\frac{(x+2)^2}{2x}\ge - \frac{x}{16}$ $$\frac{(x+2)^4}{x^3}-\frac{(x+2)^2}{2x}\ge - \frac{x}{16}$$$$\implies \frac{(2x+4)^4}{x^3}-\frac{(4x+8)^2}{2x}+x\geq 0$$$$\implies \frac{2(2(x+2))^4-(x)^2(4(x+2))^2+2(x)^4}{2(x)^3}\geq 0$$$$\implies \frac{(x+4)^2(x+\frac{8}{6})^2}{x^3}\geq 0$$$$\implies x\in (0,\infty)\cup \{\frac{-8}{6},-4\}$$