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Let $a,b,c,d,e\in (0,1)$ and such $$a+b+c+d+e=1$$ find the maximun of the value $$S=ab+ac+ad+ae+bc+bd+be+cd+ce+de$$

I Conjecture the maximun is $\dfrac{2}{5}?$,such $a=b=c=d=e=\dfrac{1}{5}$,so $$S\le\dfrac{10}{25}=\dfrac{2}{5}?$$ right?Thanks,

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  • $\begingroup$ Note that $a, b, c, d, e$ need not be restricted to $(0, 1)$. As long as the sum is $1$ and the numbers real, your conjecture holds. $\endgroup$
    – Macavity
    Commented Jul 25, 2016 at 10:04

3 Answers 3

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$$1=(a+b+c+d+e)^2 \geqslant \sum_{cyc} a^2 + 2S$$

Further from CS inequality $$(1+1+1+1+1) \cdot\sum_{cyc} a^2 \geqslant \left(\sum_{cyc} a \right)^2=1$$

Thus $1 \geqslant \frac15+2S \implies S \leqslant \frac25 $

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Since $$4(a+b+c+d+e)^2-10(ab+ac+ad+ae+bc+bd+ce+cd+ce+de)=$$ $$=(a-b)^2+(a-c)^2+(a-d)^2+(a-e)^2+(b-c)^2+(b-d)^2+(b-e)^2+(c-d)^2+(c-e)^2+(d-e)^2\geq0$$ so the answer is $\frac{2}{5}$.

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  • $\begingroup$ Nice,so for any $a_{1}+a_{2}+\cdots+a_{n}=1$,then we have $$\sum_{sym}a_{1}a_{2}\le \dfrac{\binom{n}{2}}{n^2}=\dfrac{n-1}{2n}$$? $\endgroup$
    – math110
    Commented Jul 25, 2016 at 9:59
  • $\begingroup$ Not exactly! $\sum\limits_{sym}a_1a_2=2(n-2)!\sum\limits_{1\leq i<j\leq n}a_ia_j$ $\endgroup$ Commented Jul 25, 2016 at 10:03
  • $\begingroup$ @Michale,oh,Thanks $\endgroup$
    – math110
    Commented Jul 25, 2016 at 10:06
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Let $u = \left( a , a, a , a , b, b, b , c, c, d \right)$ and $v = (b,c,d,e,c,d,e,d,e,e)$. Then by Cauchy-Schwartz \begin{align} a(b+c+d+e)+b(c+d+e)+c(d+e)+de=\langle u,v \rangle \le \|u\| \| v\| \end{align} with equality if and only if $v = \lambda u$, since $v \neq 0$ and $u\neq 0$.
So in particular the maximum is when $a = \lambda b$, $a = \lambda c$, $a = \lambda d$ and $a = \lambda e$ from the first 4 coordinates.
From the fifth coordinate we have $b = \lambda c = \lambda^2 b$, so $\lambda \in \{-1,1\}$.
As $a,b,c,d,e \ge 0$, we must have $\lambda = 1$, and thus the maximum is indeed when $a=b=c=d=e = \frac{1}{5}$.

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