Minimum value#3

$a^{2} + b^{2} + c^{2} = 1$

$\Rightarrow (a+c)^{2} + b^{2} - 1 = 2ac$

Let $S = ab + bc + 2ac$

$\Rightarrow S = b(a+c) + (a+c)^{2} + b^{2} - 1$

$\Rightarrow S = \left( b + \dfrac{a+c}{2} \right)^{2} + \dfrac{3}{4}(a+c)^{2} - 1$

$\Rightarrow S \geq -1$

$\Rightarrow$ The minimum value of $S$ is $-1$ and it happens when $b + \dfrac{a+c}{2} = 0$ and $a + c = 0$ , or $b = 0, a = -c = \pm \dfrac{1}{\sqrt{2}}$