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?

Jensen's Inequality gives
$$
\left[\frac13(a+b+c)\right]^2\le\frac13(a^2+b^2+c^2)
$$
and we know that
$$
\begin{align}
ab+bc+ca
&=\frac12\left[(a+b+c)^2-(a^2+b^2+c^2)\right]\\
&\le\frac12\left[(a+b+c)^2-\frac13(a+b+c)^2\right]\\
&=\frac13(a+b+c)^2
\end{align}
$$
and equality can be achieved when $a=b=c$. Therefore, if $a+b+c=4$, the maximum of $ab+bc+ca$ is $\frac{16}{3}$ which can be achieved if $a=b=c=\frac43$.
