<s>[INST] 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?
 [/INST] We have to use the word "triangle" at some point.
$$ a^2+b^2+c^2 \le a(b+c)+b(a+c)+c(a+b)=2(ab+ac+bc)=6$$
hence
$$(a+b+c)^2\le 12$$
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