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.
 A: Write
$$\frac{2a^2+b^2+b^2+2c^2}{2}=\frac{2a^2+b^2}{2}+\frac{b^2+2c^2}{2}\geq \sqrt{2}ab+\sqrt{2}bc$$
A: 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
A: Let $\dfrac{a^2+b^2+c^2}{ab+bc}=k>0$ as $a,b,c>0$
$\iff b^2-kb(a+c)+a^2+c^2=0$ which is a Quadratic Equation in $b$
As $b$ is real, the discriminant must be $\ge0$
i.e., $$k^2(a+c)^2-4(a^2+c^2)\ge0\iff k^2\ge\dfrac{4(a^2+c^2)}{(a+c)^2}$$
the equality occurs if $a=\dfrac{k(a+c)}2$
Now $2(a^2+c^2)-(a+c)^2=(a-c)^2\ge0\implies2(a^2+c^2)\ge(a+c)^2$
$\implies k^2\ge2$
the equality occurs if $a=c$
A: 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 $(1, \sqrt  2, 1)$.
