Timeline for An inequality related to Power sum and elementary symmetric polynomial and majorizes
Current License: CC BY-SA 4.0
16 events
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| Jun 17, 2018 at 17:04 | history | edited | Đào Thanh Oai | CC BY-SA 4.0 |
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| Jun 16, 2018 at 22:13 | comment | added | darij grinberg | Well, this inequality follows by subtracting a multiple of your first inequality from your second. | |
| Jun 16, 2018 at 19:16 | comment | added | Đào Thanh Oai | What do You think for the inequality $\frac{C_n^k}{n}(p_k(x_1,\dots,x_n) - p_k(y_1,\dots,y_n)) \ge e_k(x_1, \ldots, x_n) - e_k(y_1, \ldots, y_n)$ is true? | |
| Jun 16, 2018 at 19:08 | comment | added | Đào Thanh Oai | @darijgrinberg Thank You very very much for your helping | |
| Jun 16, 2018 at 18:28 | comment | added | darij grinberg | The "if" part fails for $n = 3$ and $\left(x_1,x_2,x_3\right) = \left(9,2,2\right)$ and $\left(y_1,y_2,y_3\right) = \left(6,6,1\right)$. | |
| Jun 16, 2018 at 17:56 | history | edited | darij grinberg | CC BY-SA 4.0 |
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| Jun 16, 2018 at 17:55 | comment | added | darij grinberg | Oh, the "if" part? I didn't notice that you even asked it; I am not sure if I really believe it, though. My proof was for the "only if" part. | |
| Jun 16, 2018 at 17:55 | comment | added | darij grinberg | Converse of what? | |
| Jun 16, 2018 at 17:53 | comment | added | Đào Thanh Oai | @darijgrinberg May You give a proof of the converse? | |
| Jun 16, 2018 at 17:46 | comment | added | darij grinberg | ... inequality $\left(1+tx_1\right)\left(1+tx_2\right)\cdots\left(1+tx_n\right) \leq \left(1+ty_1\right)\left(1+ty_2\right)\cdots\left(1+ty_n\right)$ coefficientwise. Comparing coefficients in front of $t^k$, you conclude that $e_k \left(x_1, x_2, \ldots, x_n\right) \leq e_k \left(y_1, y_2, \ldots, y_n\right)$ for all $k \geq 0$. This is probably a known trick. | |
| Jun 16, 2018 at 17:45 | comment | added | darij grinberg | This function $f : \mathbb{R}_+ \to A$ is easily seen to be log-concave (indeed, if $a$ and $b$ are two nonnegative reals and $c$ and $d$ are two elements of the interval $\left[a,b\right]$ satisfying $c+d = a+b$, then $f\left(c\right) f\left(d\right) - f\left(a\right) f\left(b\right) = \left(1+tc\right)\left(1+td\right) - \left(1+ta\right)\left(1+tb\right) = t^2\left(cd-ab\right) \geq 0$ due to the fact that $cd-ab = \left(b-c\right)\left(c-a\right) \geq 0$). So you get the ... | |
| Jun 16, 2018 at 17:42 | comment | added | darij grinberg | ... the well-known fact that $\left(y_1, y_2, \ldots, y_n\right)$ can be obtained from $\left(x_1, x_2, \ldots, x_n\right)$ by a finite sequence of steps, where each step "moves" two entries of the $n$-tuple closer to each other while preserving their sum.) Now, apply this analogue to $A = \left(\mathbb{R}_+\left[t\right], \cdot\right)$ (the multiplicative monoid of all polynomials in one variable $t$ with nonnegative real coefficients, with coefficientwise ordering) and $f\left(x\right) = 1+tx$. [Correction: When I said "group" in the above comments, I meant "monoid".] ... | |
| Jun 16, 2018 at 17:38 | comment | added | darij grinberg | ... an analogue (and generalization, if you think about it properly) of Karamata's inequality says that if an $n$-tuple $\left(x_1, x_2, \ldots, x_n\right)$ of nonnegative reals majorizes another such $n$-tuple $\left(y_1, y_2, \ldots, y_n\right)$, then $f\left(x_1\right) f\left(x_2\right) \cdots f\left(x_n\right) \leq f\left(y_1\right) f\left(y_2\right) \cdots f\left(y_n\right)$ for any partially ordered abelian group $A$ and any log-concave function $f : \mathbb{R}_+ \to A$. (This can be proven easily by recalling ... | |
| Jun 16, 2018 at 17:36 | comment | added | darij grinberg | As for the $e_k$ inequality, I think the best way is to generalize. If $A$ is a partially ordered abelian group (written multiplicatively), then we say that a function $f : \mathbb{R}_+ \to A$ is log-concave if and only if it satisfies $f\left(a\right) f\left(b\right) \leq f\left(c\right) f\left(d\right)$ whenever $a$ and $b$ are two nonnegative reals and $c$ and $d$ are two elements of the interval $\left[a,b\right]$ satisfying $c + d = a + b$. Now, ... | |
| Jun 16, 2018 at 17:32 | comment | added | darij grinberg | The $p_k$ inequality follows from Karamata's inequality, since the function $x \mapsto x^k$ is convex on $\mathbb{R}_+$. | |
| Jun 16, 2018 at 16:36 | history | asked | Đào Thanh Oai | CC BY-SA 4.0 |