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) $?
 A: 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 $1$.
A: See Jasper Loy's comment for the direct answer to your question.
To see why the minimum value is 1, use your inequality:
$$
a+b+c\ge 1.
$$
The inequality does not tell you that the minimum value is 1; for all you know at this point $a+b+c$ may always be bigger than 2; the above inequality will still be true.
One way to show that the minimum value of $a+b+c$ is 1 is to find explicit values of $a$, $b$, and $c$ with $a+b+c=1$ that satisfy the requirements set in the problem.
$a=b=c={1\over3}$ does the job.
A: Answer is 1
just expand $(a+b+c)^2$
and use the AM-GM inequality for $a^2+b^2+c^2$
so $a^2+b^2+c^2$ has a minimum value of $2(abc)^{3/2}$  which is $1/3$
and $2(ab+bc+ca)$ has a minimum value of $2*1/3=2/3$
adding both you will get 1, which is minimum value of $(a+b+c)^2$ and hence $(a+b+c)$
