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.

You can do the following:
$$\frac{a^2+b^2+c^2}{b(a+c)}\geq \frac{\frac{1}{2}(a+c)^2+b^2}{b(a+c)}=\frac{1}{2}\frac{a+c}{b}+\frac{b}{a+c}$$
where at the first step we used $2(a^2+c^2)\geq (a+c)^2$
If you now set $\frac{a+c}{b}=x$ then you just have to minimize $\frac{x}{2}+\frac{1}{x}$ over the positive real numbers.
But you have that by AM-GM $\frac{x}{2}+\frac{1}{x}\geq \sqrt{2}$ with equality if $x=\sqrt{2}$, i.e. $a+c=\sqrt{2}b$
Going back even further we want $a=c$ from the first step
