What is the minimum possible value of $(a+b+c)$? $a, b$ and $c$ are real positive numbers satisfying $ \frac 13 \le ab+bc+ca \le 1 $ and $abc \ge \frac 1{27}$ then what is the minimum possible value of $(a+b+c)$?
Applying AM $\ge$ GM gives $(a+b+c) \ge 1$ and if we apply AM $\ge$ HM gives $(a+b+c) \ge\frac 13$ but apparently $1$ is the answer, so my question is why are we not taking the second one as minimum, (since $ \frac13 \lt 1) $?

As you observed, AM-GM says that, $a+b+c\ge 1$.
*In addition, AM-GM says that equality is attained precisely when $abc=1/27$ and  $a=b=c=1/3$. 
The inequalities $1/3 \le ab+bc+ca\le 1$ are, luckily, satisfied by this choice of $a,b,c$. So the minimum achievable value of $a+b+c$ is indeed $1$.
