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Substituting $b=2-\tfrac12(a+c)$ , we have $ab+ac+bc = 4 - \tfrac12(a-2)^2 - \tfrac12(c-2)^2$ so the maximum value of $ab + ac + bc$ , subject to $a+2b+c=4$ , is $\boxed{4}$ , achieved when $a=c=2$ , $b=0$ .

$a+2b+c=4$

$(a+b)+(b+c)=4$

now using $AM\geq GM$

$\frac{(a+b)+(b+c)}{2}$ $\geq$ $\sqrt{(a+b)(b+c)}$

$2\geq \sqrt{(ab+bc+ac+b^2)}$

squaring both sides, we get

$4\geq (ab+bc+ac+b^2)$ $\Rightarrow$ $4-b^2\geq (ab+bc+ac)$

$b^2\geq 0$ $\Rightarrow$ $4-b^2\leq 4$

thus the maximum value of $(ab+bc+ac)=4$ and occurs at $b=0$ and $a=c=2$

$ab+bc+ca=b(a+c)+ca=b(4-2b)+ca \leq b(4-2b)+\left(\dfrac{c+a}{2}\right)^2=b(4-2b)+\left(\dfrac{4-2b}{2}\right)^2=4-b^2\\ b^2\geq 0\implies 4-b^2\leq 4\\ \implies\boxed{ab+bc+ca\leq 4}$ Equality occurs when $b=0,a=c=2$

Write s=a+c and s=a-c; so the above expression translates to 4 - 1/4[(s-4)^2 - d]; the following expression is maximized when s=4 and d=0. So a=c=2. Hence max value is 4.