Solution. Let's first give an elementary solution to this simple problem. Since $a, b, c \in[1,2]$, if $a \geq b \geq c$ then
$$\begin{aligned} a^{3}+2 \leq 5 a & \Leftrightarrow(a-2)\left(a^{2}+2 a-1\right) \leq 0 \\ 5 a+b^{3} \leq 5 a b+1 & \Leftrightarrow(b-1)\left(b^{2}+b+1-5 a\right) \leq 0 \\ 5 a b+c^{3} \leq 5 a b c+1 & \Leftrightarrow(c-1)\left(c^{2}+c+1-5 a b\right) \leq 0 \end{aligned}$$

The above estimations are correct because
$$\begin{array}{l} b^{2}+b+1 \leq a^{2}+a+1 \leq 2 a+a+1 \leq 5 a \\ c^{2}+c+1 \leq a^{2}+a+1 \leq 5 a \leq 5 a b \end{array}$$

Summing up the results (1), (2) and (3), we get the result. Equality holds if $a=2, b=$ $c=1$ and permutations.