The lengths $a,b$ and $c$ are the side lengths of a triangle that satisfy $ab+bc+ca=3$. Prove that $3≤a+b+c≤2√3$
I was able to find the minimum bound through some simple algebra:
$a+b+c=\sqrt{a^2+b^2+c^2+6}$
But, $a^2+b^2+c^2≥ab+bc+ca=3$
$\implies a+b+c≥\sqrt{3+6}$ $$3≤a+b+c$$
Alternatively we could use AM GM on $ab+bc+ca$ and then on $a^2+b^2+c^2$
After this I couldn't think of how to use the triangle inequality in conjunction with the given condition $ab+bc+ca=3$ to get an upper bound on the sum, any help would be appreciated, thanks