Prove that $3 \le a+b+c \le 2\sqrt{3}$ in a triangle 
Let $a,b,$ and $c$ be the lengths of the sides of a triangle satisfying $ab+bc+ca = 3.$ Prove that $3 \le a+b+c \le 2\sqrt{3}$. 

The idea I had was $(a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ca) = a^2+b^2+c^2+6 \geq 9$ by rearrangement. That takes care of the first inequality. How do I show the other inequality?

For the other side, we note that
\begin{align*}
(a+b+c)^2 &\ge 3 (ab + bc + ca)\\
&= 9.
\end{align*}
