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?
 A: 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
$$(a+b+c)^2\le 12$$
A: While working on a non-variational approach, I present a variational apporach.
By the given condition, $ab+bc+ca=3$, we get
$$
(b+c)\,\delta a+(c+a)\,\delta b+(a+b)\,\delta c=0\tag{1}
$$
For interior critical points maximizing $a+b+c$, we want
$$
\delta a+\delta b+\delta c=0\tag{2}
$$
To get $(2)$ for all variations that satisfy $(1)$, we get $b+c=c+a=a+b$, which means $a=b=c=1$. That is
$$
a+b+c=3\tag{3}
$$
The edge critical points will come when either $a=b+c$ or $b=c+a$ or $c=a+b$. Without loss of generality, assume $a=b+c$. $(1)$ becomes
$$
(2b+3c)\,\delta b+(3b+2c)\,\delta c=0\tag{4}
$$
and $(2)$ becomes
$$
2\delta b+2\delta c=0\tag{5}
$$
To get $(5)$ for all variations that satisfy $(4)$, we get $2b+3c=3b+2c$, which means $b=c=\sqrt{\frac35}$. That is
$$
a+b+c=4\sqrt{\frac35}\tag{6}
$$
The corner critical points come from $a=0$ or $b=0$ or $c=0$. Without loss of generality, assume $c=0$, which means $a=b=\sqrt3$. That is
$$
a+b+c=2\sqrt3\tag{7}
$$
Combining $(3)$, $(6)$, and $(7)$ gives
$$
3\le a+b+c\le2\sqrt3\tag{8}
$$
A: 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. 
A: For the other side, we note that
\begin{align*}
(a+b+c)^2 &\ge 3 (ab + bc + ca)\\
&= 9.
\end{align*}
