Solve the inequality $1+\frac{2}{x+1}\leq \frac{2}{x}$

$x^2-16\geq 0$

$5x+\frac{3}{2}\geq \frac{2}{3}$

$8-x\geq 2$

$\frac{2}{3}\cdot x-\frac{1}{2}\left(4x-1\right)\geq x+2\left(x-3\right)$

$4\left(x-1\right)^2\geq \left(2x-3\right)^2-7$

$x^2-2x+5\geq 0$

${\frac{\left(x+2\right)\left(x+2\right)}{6}}\geq {\frac{x}{3}+\frac{1}{2}}$