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?

Why is $a^2+b^2+c^2\geq 3$? 
I think you need to apply somewhere that you are dealing with a triangle. 
Maybe something like this?
If you sum up the following inequalities, 
$$ab+bc=b(a+c)> b^2$$
$$ab+ac=a(b+c)> a^2$$
$$ac+bc=c(a+b)> c^2$$
you obtain $a^2+b^2+c^2\leq 6$ which by your argument gives you 
$(a+b+c)^2<12.$ This implies $a+b+c\leq 2< 3.$ 
However, the left-hand side of the inequality follows from Cauchy-Schwarz inequality for triples $(a,b,c)$ and $(b,c,a)$  and
$$ab+bc+ca\leq \sqrt{a^2+b^2+c^2} \cdot \sqrt{b^2+c^2+a^2}=a^2+b^2+c^2.$$
This finally implies $a^2+b^2+c^2\geq 3$ as you assumed before. 
