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Let $x_{1},x_{2},\cdots,x_{n}\in [0,1]$, show that $$\sum_{i=1}^{n}\left[\left(\prod_{j\neq i }x_{j}\right)^{\frac{n}{2}}\cdot\sqrt{1-x_{i}}\right]\le 1$$

Maybe it can use McLaughlin inequality to solve it?I found this inequality simaler this problem

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    $\begingroup$ What makes you think this is true? $\endgroup$ – Gregory May 1 '17 at 15:40
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    $\begingroup$ According to a small python script and some simulations, this seems to hold. This proves nothing though. $\endgroup$ – Zubzub May 3 '17 at 14:43
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    $\begingroup$ A simple rewrite is $\sum_{i=1}^n{x_i^{-n/2}\sqrt{1-x_i}}\le\prod_{i=1}^n{x_i^{-n/2}}$ $\endgroup$ – didgogns May 3 '17 at 15:04
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Claim: For all integer $n$, positive $c$ satisfying $n/2 \le c$ and $x_{1},x_{2},\cdots,x_{n}\in [0,1]$,$$\sum_{i=1}^{n}\left[\left(\prod_{j\neq i }x_{j}\right)^c\cdot\sqrt{1-x_{i}}\right]\le 1$$

Proof:

Let $f(x_1, x_2, \cdots, x_n, c)=\sum_{i=1}^{n}\left[\left(\prod_{j\neq i }x_{j}\right)^c\cdot\sqrt{1-x_{i}}\right]$.

Let's proceed by mathematical induction.

Base case is trivial because $\sqrt{1-x_1} \le 1$.

For induction case, we need to check the critical points and when any of $x_i$ is $0$ or $1$.

Firstly, the cases where any of $x_i$ is $1$ is covered by induction hypothesis. Also, when $x_i=0$, $f=\left(\prod_{j\neq i }x_{j}\right)^c\le1$.

Let's compute the critical points of $f$. First note that $\frac{\partial f}{\partial x_i}=-\frac{1}{2\sqrt{1-x_i}}\left(\prod_{j\neq i }x_{j}\right)^c+\frac{c}{x_i}\sum_{k\neq i}\left[\left(\prod_{j\neq k }x_{j}\right)^c\cdot\sqrt{1-x_{k}}\right]$ and assuming $\prod_ix_i\neq 0$, $\frac{\partial f}{\partial x_i}=0$ is equivalent to$$\frac{2c-(2c-1)x_i}{2x_i^{c}\sqrt{1-x_i}}=c\sum_{k}\left[\frac{\sqrt{1-x_{k}}}{x_k^c}\right]$$and this means $\frac{2c-(2c-1)x_i}{2x_i^{c}\sqrt{1-x_i}}$ is same for all $x_i$. One can check that the function $g(x)=\frac{2c-(2c-1)x}{2x^{c}\sqrt{1-x}}$ has one local minimum, therefore on a critical point, there are only $2$ values that $x_i$s can get. Let $y, z$ be numbers satisfying $y \le z, g(y)=g(z)$ and suppose there are $l$ $x_i$s satisfying $x_i=y$, and other $n-l$ $x_i$s satisfies $x_i=z$. Then,

$f=\sum_{i=1}^{n}\left[\left(\prod_{j\neq i }x_{j}\right)^c\cdot\sqrt{1-x_{i}}\right]=y^{cl}z^{c(n-l)}\left(l\frac{\sqrt{1-y}}{y^c}+(n-l)\frac{\sqrt{1-z}}{z^c}\right)$

From $g(y)=g(z)$, $\frac{\sqrt{1-y}}{y^c}=\frac{\sqrt{1-z}}{z^c}\left(\frac{2c-(2c-1)z}{1-z}\right)\left(\frac{1-y}{2c-(2c-1)y}\right)$. Applying this,$$f=y^{cl}z^{c(n-l-1)}\sqrt{1-z}\left(l\left(\frac{2c-(2c-1)z}{1-z}\right)\left(\frac{1-y}{2c-(2c-1)y}\right)+n-l\right)$$ Note that $l\left(\frac{2c-(2c-1)z}{1-z}\right)\left(\frac{1-y}{2c-(2c-1)y}\right)+n-l\le n$ and $y^{cl}z^{c(n-l-1)}\sqrt{1-z} \le z^{c(n-1)}\sqrt{1-z} \le z^{n(n-1)/2}\sqrt{1-z}$. $z^{n(n-1)/2}\sqrt{1-z}$ is maximized at $z=1-\frac{1}{n^2-n+1}$. Then we know$$z^{c(n-1)}\sqrt{1-z} \le \left(1-\frac{1}{n^2-n+1}\right)^{n(n-1)/2}\sqrt{\frac{1}{n^2-n+1}}$$ and $\left(1-\frac{1}{n^2-n+1}\right)^{n(n-1)/2}\approx\exp(-0.5)$, in fact, it is less than $2/3$ for all integer $n>1$. Also, $\frac{2n}{3(n-0.5)}<1$ if $n \ge 2$. Therefore, we finally get$$f \le (z^{c(n-1)}\sqrt{1-z})n\le \left(1-\frac{1}{n^2-n+1}\right)^{n(n-1)/2}\sqrt{\frac{n^2}{n^2-n+1}}\le\frac{2n}{3(n-0.5)}<1$$ And we proved the claim.

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