quadratic

\hspace{-16}a+2b+c=4\Leftrightarrow \left(a+c\right)=\left(4-2b\right)$\\\\ Now Let $K=ab+bc+ca=b\left(a+c\right)+ca=b.\left(4-2b\right)+ca$\\\\ If $a,c>0$,Then $\frac{a+c}{2}\geq \sqrt{ac}\Leftrightarrow \frac{4-2b}{2}\geq \sqrt{ac}\Leftrightarrow ac\leq \left(2-b\right)^2$\\\\ So $K=b.\left(4-2b\right)+ca\leq b.\left(4-2b\right)+\left(2-b\right)^2\leq 4-b^2\leq 4$\\\\ So $\boxed{\boxed{K_{Max.}=\left(ab+bc+ca\right)_{Max.}\leq 4}}$

Thank you..

also, a + 2b + c = 4 => (a + b) + (b + c) = 4

quadratic equation with (a + b) and (b + c) as roots is given by,

x² - 4x + (ab + ac + b² + bc) = 0

Since, a, b and c are real so,

D ≥ 0 => 16 - 4(ab + ac + b² + bc) ≥ 0

=> ab + bc + ac + b² ≤ 4

=> ab + bc + ac ≤ 4 - b²

since, b is real so min(b²) = 0

Thus, max(ab + bc + ac) = 4 Ans..!!

@man111 - but if a,c are not greater than 0 then?