How find the of minimum $ab+bc+\frac{\sqrt{2}}2ac$ for $a^2+b^2=4,b^2+c^2=8$ let $a,b,c$ are real numbers and such $a^2+b^2=4,b^2+c^2=8$, find the minimum of
$$ab+bc+\dfrac{\sqrt{2}}{2}ac$$
if this problem ask find the maximum 
we  can use $AM-GM$
$$ab+bc+\dfrac{\sqrt{2}}{2}ac=\dfrac{1}{Ax}Aa\cdot xb+\dfrac{1}{Bz}Bb\cdot zc+\dfrac{\sqrt{2}}{2Cv}va\cdot Cc\le \dfrac{1}{Ax}(A^2a^2+x^2b^2)+\dfrac{1}{Bz}(B^2b^2+z^2c^2)+\dfrac{\sqrt{2}}{2Cv}(v^2a^2+C^2c^2)$$
But  for mimimum I can't solve it.Thank you 
for mimimum  I have some idea
let $$a=2\cos{x},b=2\sin{x},b=2\sqrt{2}\cos{y},c=2\sqrt{2}\sin{y}$$
and $\sin{x}=\sqrt{2}\cos{y}$
then $$ab+bc+\dfrac{\sqrt{2}}{2}ac=4\sin{x}\cos{x}+4\sqrt{2}\sin{x}\cos{y}+4\sin{y}\cos{x}$$
my nice methods
$$ab+bc+\dfrac{\sqrt{2}}{2}ac=\dfrac{\sqrt{2}}{4}(a+\sqrt{2}b+c)^2-3\sqrt{2}$$

Looks like a job for Lagrange multipliers...
let $$F(a,b,c,\lambda,\mu) = ab + bc + \frac{\sqrt{2}}{2} ac + \lambda (a^2 + b^2 - 4) + \mu(b^2 + c^2 - 8)$$
and solve $\dfrac{\partial F}{\partial a} = \dfrac{\partial F}{\partial b} = \dfrac{\partial F}{\partial c} = \dfrac{\partial F}{\partial \lambda} = \dfrac{\partial F}{\partial \mu} = 0$
