Maximum value of $ab+bc+ca$ given that $a+2b+c=4$ Question: 

Find the maximum value of $ab+bc+ca$ from the equation,
  $$a+2b+c=4$$  

My method: 

I tried making quadratic equation(in $b$) and then putting discriminant greater than or equal to $0$. It doesn't help as it yields a value greater than the answer.

Thanks in advance for the solution.

Without using calculus: Substituting $c=4-2b-a$, we get $$ab+bc+ca=ab+(a+b)(4-2b-a)=(4(a+b)-(a+b)^2)-b^2$$ and since $f(x)=4x-x^2=4-(x-2)^2$ has maximum at $(2,4)$, substituting $x:=a+b$ gives $$ab+bc+ca\le4-b^2\le4.$$
