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) $?

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 $(a+b+c)$
