Evaluating $\max(ab+bc+ac)$ 
Let a,b,c be real numbers such that $a+2b+c=4$. What is the value of $\max(ab+bc+ac)$

My attempt:
Squaring both the sides: 
$a^2 +4b^2+c^2+2ac+4bc+4ab=16$
Then I tried factoring after bringing 16 to LHS but couldn't. It's not even a quadratic in one variable or else I could have directly found the maximum. 
How do I proceed? 

From the given condition $\,2b=4-a-c\,$, then:
$$\require{cancel}
\begin{align}
2(ab+bc+ca) &= 2b(a+c)+2ac \\
 &= (a+c)(4-a-c)+2ac= \\
 &= 4a - a^2 - \bcancel{ac} + 4c - \bcancel{ac} - c^2 + \bcancel{2ac} = \\
 &= \color{red}{8} -a^2 + 4a \color{red}{-4} - c^2 +4c \color{red}{-4} = \\
 &= 8 - (a-2)^2 - (c-2)^2 \\
 &\le 8
\end{align}
$$
Equality is attained for $a=c=2\,$, so the upper bound is in fact a maximum.
