# 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?

• – lab bhattacharjee Sep 30 '14 at 3:06
• So you have $4(a+b+c)=4$ and you want to maximize $2(ab+bc+ca)$. That should give you $\frac23$ in the end. – Macavity Sep 30 '14 at 4:04
• James as Macavity pointed out, the actual expression is $4(a+b+c)=4$. Draw a diagram, you will see that there are 4 sides length a, b and c each. – Sherlock Holmes Sep 30 '14 at 6:54

@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 $\frac{2}{3}$

• How do you get $a^2+b^2+c^2\ge ab+ac+bc$? – robjohn Oct 2 '14 at 20:05
• @robjohn I have edited it in. – Sherlock Holmes Oct 3 '14 at 11:35
• I hope you don't mind. I added some to your answer to make it easier to follow. Furthermore, there is no real reason that $a,b,c\gt0$. – robjohn Oct 3 '14 at 14:38
• @robjohn "The context of the question is maximising the surface area of a rectangular prism", so shouldn't $a,b,c>0$, or, at the very least, $a,b,c \geq 0$? – Sherlock Holmes Oct 4 '14 at 1:49
• The question asked does not mention surface areas, only the motivation for the question asked. Another interesting bound is $$-\frac12(a^2+b^2+c^2)\le ab+bc+ca\le a^2+b^2+c^2$$ – robjohn Oct 4 '14 at 3:41

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 $a=b=c=\frac43$.