Given three real numbers $a,b,c$ so that $\{a,b,c\}\subset [1,2]$ . How to make $\sqrt[3]{\frac{a+b+c}{\max\{a,b,c\}}}\leq\frac{3}{2}$ a lot stronger?

Given three real numbers $a, b, c$ so that $\{a, b, c\}\subset [1, 2]$ . Easily prove that $$\sqrt[3]{\frac{a+ b+ c}{\max\{a, b, c\}}}\leq \frac{3}{2}$$

Because $$(\!a+ b+ c\!)\cdot \left (\!\frac{1}{a}+ \frac{1}{b}+ \frac{1}{c}\!\right )\leq \frac{81}{8}\Rightarrow \frac{\max\{a, b, c\}}{a+ b+ c}\geq \frac{a\cdot \frac{1}{a}+ b\cdot \frac{1}{b}+ c\cdot \frac{1}{c}}{\frac{1}{a}+ \frac{1}{b}+ \frac{1}{c}}\cdot \frac{1}{a+ b+ c}\geq \frac{8}{27}$$

How to make my own inequality a lot stronger ? Thanks for all your nice comments and problems.

$a+b+c \le 3\max \{a,b,c\},$ hence

$$\frac{a+ b+ c}{\max\{a, b, c\}} \le 3 < \frac{27}{8}.$$

Remarks:

1. we only need that $a,b,c >0$.

2. $a,b,c \in [1,2]$ is superfluous.

3. the number $3$ is the best upper bound, as $a=b=c$ show.