Given $a+b+c=4$ find $\max(ab+ac+bc)$ $a+b+c = 4$.
What is the maximum value of $ab+ac+bc$?
Could this be solved by a simple application of Jensen's inequality? If so, I am unsure what to choose for $f(x)$. If $ab+ac+bc$ is treated as a function of $a$ there seems no easy way to express $bc$ in terms of $a$.
EDIT: The context of the question is maximising the surface area of a rectangular prism.  Also I might have misinterpreted the question, because it says "the sum of the length of the edges (side lengths are a,b,c) is 4", and gives the options $\frac1{3}, \frac{2}{3}, 1, \frac{4}{3}$.
Otherwise, how would this be done?

@James, first we square both sides:
$$(a+b+c)^2=1 \implies 1=a^2+b^2+c^2+2(ab+bc+ac)\tag{1}$$
$$\because a^2+b^2+c^2 \geq ab +ac + bc\tag{2}$$
For a,b,c>0
$$ \\ 1 \geq 3(ab+bc+ac) \tag{3}$$ $$ \frac{1}{3} \geq ab +bc+ca\tag{4} \\ \square$$
$(2)$ can be easily proven by considering: $(a-b)^2 \geq 0 \\ (b-c)^2 \geq 0 \\ (a-c)^2 \geq 0 $, and adding the 3 inequalities and rearranging. 
$(3)$ follows by using $(2)$ to replace $a^2+b^2+c^2$ in $(1)$ with the smaller $ab+ac+bc$.
Surface area is $2(ab+bc+ca)$, so maximum surface area is simply $\frac{2}{3}$
