6

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 $ab + bc + cd + de + ef$?

leonbloy
  • 63,430
user58452
  • 61
  • 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 04:46
  • @gnometorule agreed. – Rustyn Jan 17 '13 at 04:48
  • Yes the Answer is 1/4..still me a bit confused – user58452 Jan 17 '13 at 05:26
  • 1
    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 05:28
  • 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 05:34
  • @lyj...okay i am getting you...except for how can you say maximum occurs when 2nd expression is zero..sorry for such newbie things – user58452 Jan 17 '13 at 05:49
  • @user58452 Achieving maximum $1/4$ is just computational once we show the inequality as lyj did. – user123454321 Jan 17 '13 at 06:08
  • Have you tried Lagrange multiplier technique? – Mhenni Benghorbal Jan 17 '13 at 06:18
  • @gilyoungcheong you are correct in your reasoning, i retract my answer. – Rustyn Jan 17 '13 at 06:20
  • @RustynYazdanpour Solution was way easier than what I have been thinking... – user123454321 Jan 17 '13 at 06:22

3 Answers3

1

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.

user123454321
  • 2,428
0

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 $1/4$.

timidpueo
  • 2,049
0

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 $a = 0,\, b = \frac{1}{3},\, c = \frac{1}{2},\, d = \frac{1}{6}, e = 0,\, f = 0.$

cats
  • 4,298