Seems like $uvw$

@Pi Han Goh This is the solution.

Let $x=\frac { \sqrt { 3 } +1 }{ 2\sqrt { 3 } } ,y=\frac { \sqrt { 3 } -1 }{ 2\sqrt { 3 } } , z=0$ , we obtain the value of 12. (My most ridiculous part is this one... I mean, how would a random person know that these numbers are optimal???)

Now let's prove it's the minimum.

Let $p=x+y+z=1, q=xy+yz+zx, r=xyz$ .

$q\ge 12(p({ p }^{ 2 }-3q)+3r)({ q }^{ 2 }-2pr)$

$=>q\ge 12(1-3q+3r)({ q }^{ 2 }-2r)$

Note that $pq \ge 9r$ $=>12(1-3q+3r)({ q }^{ 2 }-2r) \le 12{ q }^{ 2 }(1-3q+\frac { { q } }{ 3 } )=4{ q }^{ 2 }(3=8q)$ ( $xyz \ge 0$ )

It is obviously shown that $4q(3-8q) \le 1 => (1-4q)(1-8q) \ge 0$ for all $q \ge \frac { 1 }{ 4 }$ , which leaves us to prove that the inequality above is true for $0 \le q \le \frac { 1 }{ 4 }$ .

Back to the original inequality, we can see that $q\ge 12{ q }^{ 2 }(1-3q)\ge 12{ q }^{ 2 }(1-3q)+12r(3{ q }^{ 2 }+6q-2)-72{ r }^{ 2 }$ , as $3{ q }^{ 2 }+6q-2$ and $-72r^2$ are all non-positive.

It remains to prove that $q\ge 12{ q }^{ 2 }(1-3q)$ which is obvious by AM-GM:

$4*3q(1-3q) \le (3q+1-3q)^2 =1$ .

I didn't have much time to write this solution, so if there is anything not-so-easy to understand, just leave a comment here and I'll try to explain.