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$x_1, x_2, x_3, ... ,x_m > 0 \quad \forall m\geq 3$ ,

$x_1+x_2+x_3+...+x_m = 1$.

What is maximum of

$x_1 \cdot x_2\cdot x_3\cdot ... \cdot x_m\left(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}+\ldots + \frac{1}{x_m} - 1\right)$ ?

For $m=3$, i got $\frac{8}{27}$, but i can't find the answer for m>3 :( . Thanks

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  • $\begingroup$ I suppose one could conjecture that $x_1=x_2=\cdots=x_m=1/m$ gives the maximum, which is: $$\frac{1}{m^{m-2}}-\frac{1}{m^m} = \frac{m^2-1}{m^m}$$ and then set out to prove it on a case-by-case basis (this agrees with $m=1,2,3,4$ at least), but I don't see a simple way to prove the general case. $\endgroup$ – Nicholas Stull Apr 18 '16 at 17:07
  • $\begingroup$ @NicholasStull nice help ^_^ , thank you very much $\endgroup$ – Makoto Daiwa Ambara Apr 19 '16 at 1:15
  • $\begingroup$ Apologies that it isn't more elegant, but I couldn't come up with anything that wasn't just a brute-force solution to it, even using the rather nice idea that Martin R had for the $m=4$ case. In any case, you're welcome. $\endgroup$ – Nicholas Stull Apr 19 '16 at 1:18
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Let $$f(x_1,\ldots,x_m) = \left(\sum_{k=1}^m \left(\prod_{\substack{j=1\\j\neq k}}^m x_j\right)-\prod_{j=1}^m x_j\right)$$

Since $f$ is continuous (a polynomial), we hence know that $f$ attains a maximum on the compact set $$\left\{(x_1,\ldots,x_m)\in\mathbb{R}^m\mid x_1,\ldots,x_m\geq 0, \sum_{k=1}^m x_k = 1\right\}$$ at some point $(a_1,\ldots,a_m)$.

First note that at most one of $a_1,\ldots, a_m$ can be $0$, else the whole thing is $0$, and $0$ is not a maximum value of this function. If we can show that the maximum corresponds to the case $a_1=a_2=\cdots = a_m = \frac{1}{m}$, then the maximum is $\frac{m^2-1}{m^m}$ as claimed in my comment.

Let us assume $a_1=a_2=\cdots=a_m$ does not hold. WLOG, assume $a_1\neq a_2$. We then have the following:

$$f(a_1,\ldots,a_n) = a_1a_2\left[\sum_{k=3}^m \left(\prod_{\substack{j=3\\j\neq k}}^m a_j\right) - \prod_{j=3}^m a_j\right] + (a_1+a_2)\prod_{j=3}^m a_j$$

Further, we have: \begin{align} &f\left(\frac{a_1+a_2}{2},\frac{a_1+a_2}{2},a_3,\ldots,a_m\right) \\ &= \frac{(a_1+a_2)^2}{4}\left[\sum_{k=3}^m \left(\prod_{\substack{j=3\\j\neq k}}^m a_j\right) - \prod_{j=3}^m a_j\right] + (a_1+a_2)\prod_{j=3}^m a_j \end{align}

from which we deduce \begin{align} &f\left(\frac{a_1+a_2}{2},\frac{a_1+a_2}{2},a_3,\ldots,a_m\right) - f(a_1,\ldots,a_m) \\ &= \left(\frac{(a_1+a_2)^2}{4}-a_1a_2\right)\left[\sum_{k=3}^m \left(\prod_{\substack{j=3\\j\neq k}}^m a_j\right) - \prod_{j=3}^m a_j\right]\\ &= \left(\frac{(a_1-a_2)^2}{4}\right)\left[\sum_{k=3}^m \left(\prod_{\substack{j=3\\j\neq k}}^m a_j\right) - \prod_{j=3}^m a_j\right]\\ &= \left(\frac{(a_1-a_2)^2}{4}\right)\left[\sum_{k=3}^m \left((1-a_k)\prod_{\substack{j=3\\j\neq k}}^m a_j\right) \right] \end{align}

and we claim that $$\left(\frac{(a_1-a_2)^2}{4}\right)\left[\sum_{k=3}^m \left((1-a_k)\prod_{\substack{j=3\\j\neq k}}^m a_j\right) \right] > 0$$

Indeed, $(a_1-a_2)^2>0$ if $a_1\neq a_2$, and since we have already noted that at most one of $a_1,\ldots,a_m$ can be zero, then the product in the sum is nonzero (and in fact strictly positive) for at least one $k\in\{1,\ldots,m\}$. Additionally, since $0 \leq x_k < 1$, $(1-x_k) > 0$ for every $k\in\{1,\ldots,m\}$.

Hence we have $f\left(\frac{a_1+a_2}{2},\frac{a_1+a_2}{2},a_3,\ldots,a_m\right) - f(a_1,\ldots,a_m) > 0$, which means we that $f$ does not attain a maximum value at $(a_1,\ldots,a_m)$, which is a contradiction.

Hence we have that $a_1=a_2=\ldots=a_m=\frac{1}{m}$, and in this case we have a max of $$f\left(\frac{1}{m},\ldots,\frac{1}{m}\right) = \frac{m^2-1}{m^m}$$ (to see why this is, it's actually clearer to look at the original form $x_1\cdots x_m(\frac{1}{x_1}+\cdots + \frac{1}{x_m})$, with $x_1=\cdots=x_m = \frac{1}{m}$, but it is not difficult either way)


The idea behind this solution came from Martin R's answer to your question for the case $m=4$, If $a,b,c,d > 0$ and $a+b+c+d=1$ , find the maximum value of $M=abc+abd+acd+bcd-abcd$, so certainly much of the credit should go to him.

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