I have been thinking about it for several days and still have not solved it completely. This is the problem:

My formula: Outcome=$(a_1*b_1)+(a_2*b_2)+(a_3*b_3)...+(a_n*b_n),$ where $0\le a\le 1,$ $0\le b\le1, ~~a+b\le 1, ~~a_1+a_2+a_3...+a_n\le 1.$

I would like to find the maximum of this function based on the value of n. Just by trying out different numbers, I have found that the maximum should be (n-1)/n, but I have not been able to prove it mathematically yet. Any help is very much appreciated!


closed as unclear what you're asking by kjetil b halvorsen, Chris Godsil, hardmath, JonMark Perry, user223391 Oct 22 '17 at 14:47

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  • $\begingroup$ what is $a$? is it a vector? does $0\leq a \leq 1$ means it is bounded by $0$ and $1$ element wise? $\endgroup$ – Siong Thye Goh Oct 15 '17 at 17:35
  • $\begingroup$ Finding the maximum of a multivariable function (you say there are $n$ parameters) is naturally more difficult than the single-variable function cases. In this case you have some restrictions on the parameters. But you should try to clarify the problem. I'm guessing the $n$ parameters are $a_1,a_2,\ldots,a_n$ since you say that there are $n$ parameters, and then $b_i$'s would be fixed. But that's just guesswork. $\endgroup$ – hardmath Oct 16 '17 at 2:26

Since $0\leq a_i\leq 1$, we have from $a_i+b_i\leq 1$ that


Next assume $\sum_ia_i=x\in[0,1]$. Then use

$$\frac{1}{n}\sum_{i=1}^n a_i^2\geq\left(\frac{1}{n}\sum_{i=1}^na_i\right)^2=\frac{x^2}{n^2},$$

we obtain

$$\sum_{i=1}^na_ib_i\leq x-\frac{x^2}{n}.$$

The right-hand side is monotonically increasing over $x\in[0,\frac{n}{2}]$. Therefore it is bounded by its endpoint value at $x=1$ so long as $n\geq 2$. Therefore

$$\sum_{i=1}^na_ib_i\leq 1-\frac{1}{n}.$$

For all three inequalities to take equal sign, we have $b_i=1-a_i$, $a_i=\frac{x}{n}$ and $x=1$. Therefore $a_i=\frac{1}{n}$ and $b_i=\frac{n-1}{n}$. Since equality is achievable, the upper bound is the maximum.

  • $\begingroup$ $\sum_ia_i$ is supposed to be $\le 1$, not $=1$. $\endgroup$ – amsmath Oct 15 '17 at 17:45
  • $\begingroup$ I assumed $a_1+a_2+\cdots+a_n=1$. In fact the condition is $\leq 1$. You can change it to $a_1+a_2+\cdots+a_n=x\in[0,1]$, go through the same process and then try to maximize the upper bound as a function of $x$. Then you'll see the maximum is achieved at $x=1$. $\endgroup$ – Zhuoran He Oct 15 '17 at 17:46
  • $\begingroup$ But when you go through the same process with $x$, then you get the upper bound $1-x^2/n$, which is larger than $1-1/n$. $\endgroup$ – amsmath Oct 15 '17 at 17:53
  • $\begingroup$ No you should get $x-\frac{x^2}{n}$ The function is monotonically increasing for $x\leq\frac{n}{2}$. So long as $n\geq 2$, the argument holds. $\endgroup$ – Zhuoran He Oct 15 '17 at 17:54
  • $\begingroup$ The math looks prettier without the $x$. But should I really edit it? $\endgroup$ – Zhuoran He Oct 15 '17 at 17:56

Hint :

it's just an application of the inequality of Chebyshev.

You have just to remark that we have this :

$a_1\geq a_2 \cdots \geq a_n$


$b_1\leq b_2 \cdots \leq b_n$

Finally conclude .


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