Team Selection Test

Letting $a=0$ and $b=c=2$ we obtain $2\ge M$. We will show that $M=2$ works.

Without loss of generality we may assume that $\max \{a,b,c\}=c$. Let $x=a+b$ and $y=ab$. We have $c\ge \frac{a+b+c}{3}=\frac{4}{3}$ and hence $x=a+b\le \frac{8}{3}$.

Then $$a^2+b^2+c^2+3abc\ge 2(ab+bc+ca)⟺x^2-2y+(4-x{)}^2+3y(4-x)\ge 2(y+x(4-x))⟺4(x-2{)}^2+y(8-3x)\ge 0$$ follows.