Minimizing in 3 variables 
Find the least possible value of the fraction $\dfrac{a^2+b^2+c^2}{ab+bc}$, where $a,b,c > 0$.

My try:
$a^2+b^2+c^2 = (a+c)^2 - 2ac + b^2$,
$= (a+c)/b  +b/(a+c)  -2ac/b(a+c)$
AM > GM
$3\sqrt[3]{-2ac/b(a+c)}$
And I cant somehow move on.

Yet another approach: The expression does not change if $(a, b, c)$
are multiplied by a common factor, so we can assume that $a+c=2$. Then
$$
 \frac{a^2+b^2+c^2}{ab+bc} =  \frac 12 \left (\frac{a^2+(2-a)^2}{b} + b
 \right) \ge \sqrt{a^2 + (2-a)^2} = \sqrt{2(a-1)^2 + 2} \, ,
$$
using $AM \ge GM$. It follows that
$$
\frac{a^2+b^2+c^2}{ab+bc} \ge \sqrt 2 \, ,
$$
with equality if and only if $(a, b, c)$ is a multiple of $(1, \sqrt  2, 1)$.
