The Answer Is Not 3

Observe that all the expressions are homogenous, hence we may make the simplifying assumption that $a+b+c = 1$ (Otherwise, divide through out by $k = a + b + c$ , and replace the variable with $a^* = \frac{a}{k}, b^* = \frac{b}{k} , c^* = \frac{c}{k}$ .)

We have $1 = (a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab+bc+ca) = \frac{9}{2} ( ab+bc+ca)$ so $(ab+bc+ca) = \frac{ 2}{9}$ . Let $abc = M$ , which is a positive number.

Consider the cubic equation which has roots $a, b, c$ . By Vieta's, the equation is $X^3 - X^2 + \frac{ 2}{9}X - M = 0$ . Since it has 3 real (positive) roots, it has a non-negative discriminant. As such, this tells us that

$4 - 19683 M^2 \geq 0 \Rightarrow M \leq \sqrt{ \frac{ 4}{ 19683 } } .$

Thus, the minimum value of the expression would be $\frac{ a+b+c} { \sqrt[3]{abc} } = \frac{ 1} { \sqrt[3]{M} } \geq \sqrt[6]{\frac{19683} {4} } = \boxed{4.124}$

Note: By factoring $X^3 - X^2 + \frac{2}{9} X - \sqrt{ \frac{4}{19683} } = [ x -\frac{1}{9} ( 3 - \sqrt{3}) ][ x -\frac{1}{9} ( 3 - \sqrt{3}) ][ x -\frac{1}{9} ( 3 + 2 \sqrt{3}) ]$ , we see that equality occurs at $( 3 - \sqrt{3}, 3 - \sqrt{3}, 3 + 2 \sqrt{3} )$ , and permutations and constant multiples.

I finally added a solution to this series of inequality questions with strange solution sets :)
On hindsight, the equations would be slightly nicer if we used $a + b +c = 9$ .
This question was adapted from an Iranian Olympiad problem which had the condition $a^2 + b^2 + c^2 = 2( ab + bc + ca )$ , with a similar setup. With that condition, it was easier to guess the equality condition, and push through with the solution.