Inequalities

Please edit the word 'ate' in the first line of the problem..

Given $a+2b+c=4$ . By A.M-G.M, $ab+bc+ca \leq b(a+c)+\frac{(a+c)^2}{4}$ since $a+c=4-2b$ , it implies that $ab+bc+ca \leq b(4-2b)+\frac{(4-2b)^2}{4}$ now, we can simplify this as $ab+bc+ca \leq 4-b^2$ . So, we want the maximum value of the expression $4-b^2$ .For that purpose, $b^2$ has to be minimum. So, $b=0$ , and maximum value is $4-0=\boxed{4}$