# What is the Maximum Value?

If $a, b, c, d, e$ and $f$ are non negative real numbers such that $a + b + c + d + e + f = 1$, then what is the maximum value of $ab + bc + cd + de + ef$?

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I'm pretty sure this should be $\frac{1}{4}$, similarly to simply maximizing $ab$ subject to the same constraint. – gnometorule Jan 17 '13 at 4:46
@gnometorule agreed. – Rustyn Jan 17 '13 at 4:48
Yes the Answer is 1/4..still me a bit confused – user58452 Jan 17 '13 at 5:26
My Approach : a + c + e + b + d + f = 1 So maximum value of (a + c + e) . (b + d + f) is (1/2).(1/2) i.e. (ab + bc + cd + de + ef) + (ad + af + cf + be) = (1/4) and now i am stuck...!! – user58452 Jan 17 '13 at 5:28
As you said, $(a+c+e)(b+d+f) \le \frac{1}{4},$ from which $(ab + bc + cd + de + ef) + (ad + af + cf + be) \le \frac{1}{4}.$ Since all of our variables are non-negative, the second expression is also non-negative, which means that $(ab + bc + cd + de + ef) \le \frac{1}{4}$ as well. The maximum occurs when the second expression is $0$ and when $a+c+e = b+d+f,$ which is easy to achieve. – cats Jan 17 '13 at 5:34

Note that lyj's comment pretty much answers this question. On top of that, we can achieve this maximum by taking $(a, b, c, d, e, f) = (0, 0, 11/32, 1/2, 5/32, 0)$.

I.e., We finish showing the following.

1. The quantity in question has an upper bound $1/4$ (by lyj's argument).

2. The upper bound $1/4$ can be achieved.

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I'm pretty sure the answer is $1/4$, but I can't prove it rigorously...

We want to multiply the two largest numbers we can together, so simply letting $a=1/2$ and $b=1/2$ will do the trick.

In fact if we let any of $b, c, d, e$ equal $1/2$, and let the number before and after it add to $1/2$, we will also get $1/4$.

So for example we can let $e=1/2$, and also let $d=1/8$ and $f=3/8$, we will get a maximum value of $1/4$.

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Here's the full solution in case the comments weren't enough.

Note that $(a+c+e)(b+d+f) = (a+c+e)(1-(a+c+e)) \le \frac{1}{4},$ with equality iff $a + c + e = \frac{1}{2}.$ Expanding the first expression above gives $(ab+bc+cd+de+ef)+(ad+af+be+cf) \le \frac{1}{4}.$ Since all of our variables are non-negative, $ad + af + be + cf \ge 0,$ or $-(ad+af+be+cf) \le 0,$ which gives $ab+bc+cd+de+ef \le \frac{1}{4} - (ad+af+be+cf) \le \frac{1}{4}.$ To achieve equality, we need the following:

1) $a + c + e = \frac{1}{2},$ our original condition.

2) $\frac{1}{4} - (ad + af + be + cf) = \frac{1}{4},\, \textrm{ i.e. } ad + af + be + cf = 0.$

So we have shown that the upper bound is $\frac{1}{4},$ and that we can achieve this upper bound. For example, let $a = \frac{1}{2} = b$ and $c = d = e = f = 0.$ Another example is $a = 0,\, b = \frac{1}{3},\, c = \frac{1}{2},\, d = \frac{1}{6}, e = 0,\, f = 0.$

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