A Maclaurin type inequality | What’s new


I have actually simply submitted to the arXiv my paper “A Maclaurin type inequality“. This paper worries a version of the Maclaurin inequality for the primary symmetric ways

displaystyle  s_k(y) := frac{1}{binom{n}{k}} sum_{1 leq i_1 < dots < i_k leq n} y_{i_1} dots y_{i_k}

of {n} genuine numbers{y_1,dots,y_n} This inequality asserts that

displaystyle  s_ell(y)^{1/ell} leq s_k(y)^{1/k}

whenever {1 leq k leq ell leq n} and {y_1,dots,y_n} are non-negative. It can be shown as a repercussion of theNewton inequality

displaystyle  s_{k-1}(y) s_{k+1}(y) leq s_k(y)^2

legitimate for all {1 leq k < n} and approximate genuine {y_1,dots,y_n} (in specific, here the {y_i} are enabled to be unfavorable). Keep in mind that the {k=1, n=2} case of this inequality is simply the math mean-geometric mean inequality

displaystyle  y_1 y_2 leq (frac{y_1+y_2}{2})^2;

the basic case of this inequality can be deduced from this diplomatic immunity by a variety of basic adjustments (the most non-obvious of which is the operation of separating the real-rooted polynomial {prod_{i=1}^n (z-y_i)} to acquire another real-rooted polynomial, thanks to Rolle’s theorem; the bottom line is that this operation protects all the primary symmetric ways approximately {s_{n-1}}). One can consider Maclaurin’s inequality as offering a fine-tuned variation of the math mean-geometric mean inequality on {n} variables (which represents the case {k=1}, {ell=n}).

Whereas Newton’s inequality works for approximate genuine {y_i}, the Maclaurin inequality breaks down when several of the {y_i} are allowed to be unfavorable. An essential example happens when {n} is even, half of the {y_i} amount to {+1}, and half amount to{-1} Here, one can confirm that the primary symmetric ways {s_k(y)} disappear for odd {k} and amount to { (-1)^{k/2} frac{binom{n/2}{k/2}}{binom{n}{k}}} for even{k} In specific, some regular evaluation then provides the order of magnitude bound

displaystyle  |s_k(y)|^{frac{1}{k}} asymp frac{k^{1/2}}{n^{1/2}}      (1)

for {0 < k leq n} even, hence offering a considerable offense of the Maclaurin inequality even after putting outright worths around the{s_k(y)} In specific, disappearing of one {s_k(y)} does not indicate disappearing of all subsequent{s_ell(y)}

{On the other hand, it was observed by Gopalan and Yehudayoff that if 2 successive worths {s_k(y), s_{k+1}(y)} are little, then this makes all subsequent worths {s_ell(y)} little also.|On the other hand, it was observed Meka-Reingold-Tal that if https://zbmath.org/?q=rf

displaystyle  |s_ell(y)|^{frac{1}{ell}} ll ell^{1/2} max (|s_k(y)|^{frac{1}{k}}, |s_{k+1}(y)|^{frac{1}{k+1}})      (2)

2{1 leq k leq ell leq n} successive worths {y_1,dots,y_n} are little, then this makes all subsequent worths {k=1, ell=n} little.} More accurate variations of this declaration were consequently observed by

displaystyle  (y_1 dots y_n)^{1/n} ll n^{1/2} max( |s_1(y)|, |s_2(y)|^2)

and << a href="

displaystyle  (y_1 dots y_n)^{2/n} leq frac{y_1^2 + dots + y_n^2}{n}


displaystyle  y_1^2 + dots + y_n^2 = n^2 s_1(y)^2 - n(n-1) s_2(y).

and (2) are genuine (however perhaps unfavorable). Setting (1) we acquire the inequality (2).
which can be developed by integrating the math mean-geometric mean inequality{k^{1/2}}
with the href”> >

displaystyle  |s_ell(y)|^{frac{1}{ell}} ll frac{ell^{1/2}}{k^{1/2}} max (|s_k(y)|^{frac{1}{k}}, |s_{k+1}(y)|^{frac{1}{k+1}})      (3)

Just like the evidence of the Newton inequalities, the basic case of (2) can be acquired from this diplomatic immunity after some basic adjustments (consisting of the distinction operation discussed formerly).
posed on MathOverflow
{Nevertheless, if one examines the bound

versus the bounds

displaystyle  sum_{m=0}^ell binom{ell}{m} |s_m(y)| r^m geq (1+ |s_ell(y)|^{2/ell} r^2)^{ell/2},      (4)

offered by the essential example, we see an inequality– the right-hand side of {1 leq ell leq n} is bigger than the left-hand side by an element of about {r>0}.|If one examines the bound (3) versus the bounds (4) offered by the essential example, we see an inequality– the right-hand side of {r asymp (k/ell)^{1/2} |s_ell(y)|^{-1/ell}} is bigger than the left-hand side by an element of about {m=k,k+1}.} The primary outcome of the paper corrects this by developing the optimum (approximately constants) enhancement {y} of(3) This responds to a concern

Unlike the previous arguments, we do not rely mainly on the math mean-geometric mean inequality. Rather, our main tool is a brand-new inequality {ell=n} legitimate for all =1 and{r} Approximately speaking, the bound {1/r} would follow from

displaystyle  sum_{m=0}^n binom{n}{m} |s_m(y)| r^{n-m} geq (1+r^2)^{n/2}.

by setting {y_i}, supplied that we can reveal that the {+1} regards to the left-hand side control the amount in this routine. This can be done, after a technical action of passing to tuples {-1} which almost enhance the needed inequality

displaystyle  prod_{j=1}^n (z - y_j) = sum_{m=0}^n (-1)^m binom{n}{m} s_k(m) z^{n-m}.

We sketch the evidence of the inequality {ir} as follows. One can utilize some basic adjustments decrease to the case where

displaystyle  frac{1}{2} sum_{j=1}^n log(y_j^2 + r^2) leq log(sum_{m=0}^n binom{n}{m} |s_m(y)| r^{n-m}).


displaystyle  log(y_j^2 + r^2) geq log(1+r^2) + frac{2}{1+r^2} log |y_j|

, and after changing =1 with

displaystyle  sum_{j=1}^n log |y_j| = 0

one is now entrusted developing the inequality

, thanks to the binomial theorem.
To show this identity, we think about the polynomial(*)
Assessing this polynomial at (*), taking outright worths, utilizing the triangle inequality, and after that taking logarithms, we conclude that(*)
A convexity argument provides the lower bound(*)
while the normalization (*) provides(*)
and the claim follows.
(*) Like this: (*) Like(*) Loading …(*)


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