$\text{ If }f:R\to R\text{ and }g:R\to R\text{ are given by }f\left(x\right)=\left|x\right|\text{ and }g\left(x\right)=\left[x\right] \text{ for each }x\in R,\text{ then }\{x\in R:g(f\left(x\right))\le f(g\left(x\right)\left)\right\}=$

 $$\begin{gathered} f\left(x\right)=\left| x \right| and g\left(x\right) =\left[x\right] \\ \Rightarrow g\left(f\left(x\right)\right) = g\left(\left|x\right|\right) =\left[\left|x\right|\right] and f\left(g\left(x\right)\right)=f\left(\left[x\right]\right)=\left|\left[x\right]\right| \\ when x\ge 0 , \left[\left|x\right|\right]= \left[x\right]=\left|\left[x\right]\right| \end{gathered}$$

$$\begin{gathered} therefore g\left(f\left(x\right)\right)=f\left(g\left(x\right)\right) \\ when x<0 , \left[x\right]\le x<0 \left( \sin ce \left[x\right]\le x for all x\right) \\ \left|\left[x\right]\right|\ge \left|x\right|\ge \left[\left|x\right|\right] \end{gathered}$$

$$\begin{gathered} \Rightarrow f\left(g\left(x\right)\right)\ge g\left(f\left(x\right)\right) \\ therefore g\left(f\left(x \right)\right)\le f\left(g\left(x\right)\right) for all x\in R \end{gathered}$$