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I have just uploaded to the arXiv my paper “A Maclaurin type inequality“. This paper concerns a variant of the Maclaurin inequality for the elementary symmetric means
of real numbers . This inequality asserts that whenever and are non-negative. It can be proven as a consequence of the Newton inequality valid for all and arbitrary real (in particular, here the are allowed to be negative). Note that the case of this inequality is just the arithmetic mean-geometric mean inequality the general case of this inequality can be deduced from this special case by a number of standard manipulations (the most non-obvious of which is the operation of differentiating the real-rooted polynomial to obtain another real-rooted polynomial, thanks to Rolle’s theorem; the key point is that this operation preserves all the elementary symmetric means up to ). One can think of Maclaurin’s inequality as providing a refined version of the arithmetic mean-geometric mean inequality on variables (which corresponds to the case , ).Whereas Newton’s inequality works for arbitrary real , the Maclaurin inequality breaks down once one or more of the are permitted to be negative. A key example occurs when is even, half of the are equal to , and half are equal to . Here, one can verify that the elementary symmetric means vanish for odd and are equal to for even . In particular, some routine estimation then gives the order of magnitude bound
for even, thus giving a significant violation of the Maclaurin inequality even after putting absolute values around the . In particular, vanishing of one does not imply vanishing of all subsequent .On the other hand, it was observed by Gopalan and Yehudayoff that if two consecutive values are small, then this makes all subsequent values small as well. More precise versions of this statement were subsequently observed by Meka-Reingold-Tal and Doron-Hatami-Hoza, who obtained estimates of the shape
whenever and are real (but possibly negative). For instance, setting we obtain the inequality which can be established by combining the arithmetic mean-geometric mean inequality with the Newton identity As with the proof of the Newton inequalities, the general case of (2) can be obtained from this special case after some standard manipulations (including the differentiation operation mentioned previously).However, if one inspects the bound (2) against the bounds (1) given by the key example, we see a mismatch – the right-hand side of (2) is larger than the left-hand side by a factor of about . The main result of the paper rectifies this by establishing the optimal (up to constants) improvement
of (2). This answers a question posed on MathOverflow.Unlike the previous arguments, we do not rely primarily on the arithmetic mean-geometric mean inequality. Instead, our primary tool is a new inequality
valid for all and . Roughly speaking, the bound (3) would follow from (4) by setting , provided that we can show that the terms of the left-hand side dominate the sum in this regime. This can be done, after a technical step of passing to tuples which nearly optimize the required inequality (3).We sketch the proof of the inequality (4) as follows. One can use some standard manipulations reduce to the case where and , and after replacing with one is now left with establishing the inequality
Note that equality is attained in the previously discussed example with half of the equal to and the other half equal to , thanks to the binomial theorem.To prove this identity, we consider the polynomial
Evaluating this polynomial at , taking absolute values, using the triangle inequality, and then taking logarithms, we conclude that A convexity argument gives the lower bound while the normalization gives and the claim follows.
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