what's the maximum value of the algebraic expression? 
Suppose that $a,b,c>0$ and satisfy the equation
  $$
a^2+b^2+4c^2=1,
$$
  then what's the maximum of $F(a,b,c)=ab+2ac+3\sqrt{2}bc$ ?

**
My way: $F(a,b,c)=ab+\sqrt{2}bc+2(ac+\sqrt{2}bc)=\cdots$
 A: Or using CS inequality:
$$2=(a^2+\tfrac12b^2+\tfrac12b^2+4c^2)(4c^2+2a^2+4c^2+2b^2) \ge (2ca+ab+3\sqrt2bc)^2$$
with equality when $b^2=2a^2=4c^2=\frac25$.
--
P.S. The vectors used for CS are $u = (a, \frac1{\sqrt2}b, \frac1{\sqrt2}b,2c)$ and $v = (2c, \sqrt2a, 2c, \sqrt2b)$.  So $|u|=1, |v|=\sqrt2$ and $u\cdot v$ is the expression we want to maximize.
A: Use AM-GM inequality we have
$$\dfrac{\sqrt{2}}{2}a^2+\dfrac{\sqrt{2}}{4}b^2\ge ab\tag{1}$$
$$\sqrt{2}c^2+\dfrac{\sqrt{2}}{2}a^2\ge ac\tag{2}$$
$$\dfrac{3\sqrt{2}}{4}b^2+3\sqrt{2}c^2\ge 3\sqrt{2}bc\tag{3}$$
$(1)+(2)+(3)$ we have 
$$ab+ac+3\sqrt{2}bc\le \sqrt{2}(a^2+b^2+4c^2)=\sqrt{2}$$
A: I suppose that Lagrange multipliers is a simple way to get the solution. 
Consider, as usual, $$F=a b+2 a c+3 \sqrt{2} b c+\lambda  \left(a^2+b^2+4 c^2-1\right)$$ and compute the partial derivatives $$F'_a=2 a \lambda +b+2 c$$ $$F'_b=a+2 b \lambda +3 \sqrt{2} c$$ $$F'_c=2 a+3 \sqrt{2} b+8 c \lambda$$ $$F'_\lambda=a^2+b^2+4 c^2-1$$ Setting the partials to zero, eliminate $a$ for the first equation $F'_a=0$ to get $$a=\frac{-b-2 c}{2 \lambda }$$ Replace in the second equation $F'_b=0$ and eliminate $b$ to get $$b=\frac{c \left(2-6 \sqrt{2} \lambda \right)}{4 \lambda ^2-1}\implies a=\frac{c \left(3 \sqrt{2}-4 \lambda \right)}{4 \lambda ^2-1}$$ Replace in the third equation $F'_c=0$ to get $$\frac{4 c \left(8 \lambda ^3-13 \lambda +3 \sqrt{2}\right)}{4 \lambda ^2-1}=0$$ Since $c>0$, assuming $4\lambda^2 \neq 1$, $\lambda$ is then a solution of $$8 \lambda ^3-13 \lambda +3 \sqrt{2}=0$$ Using Cardano method for solving the cubic, we have three real roots $$\lambda_1=\frac{1}{2 \sqrt{2}}\quad \lambda_2=\frac{3}{2 \sqrt{2}}\quad\lambda_3=-\sqrt{2}$$ Plugging in the constraint leads to $$c^2=\frac{\left(1-4 \lambda ^2\right)^2}{8 \lambda  \left(8 \lambda ^3+7 \lambda -6
   \sqrt{2}\right)+26}$$ Replacing $\lambda$ by its possible  values leads to $$\lambda_1=\frac{1}{2 \sqrt{2}}\implies c_1=\frac{1}{2 \sqrt{10}}\implies a_1=-\frac{2}{\sqrt{5}}\implies b_1=\frac{1}{\sqrt{10}}$$ $$\lambda_2=\frac{3}{2 \sqrt{2}}\implies c_2=\frac{1}{2 \sqrt{2}}\implies a_2=0\implies b_2=-\frac{1}{\sqrt{2}}$$ $$\lambda_3=-\sqrt{2}\implies c_3=\frac{1}{\sqrt{10}}\implies a_3=\frac{1}{\sqrt{5}}\implies b_3=\sqrt{\frac{2}{5}}$$ (we ignore the negative roots of $c$).
So, only the last set gives non negative values for $a,b,c$. Plugging these numbers in the objective function $a b+2 a c+3 \sqrt{2} b c$, we then get  a maximum value equal to $\sqrt 2$.
