Almost-converses to the AM-GM inequality - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T06:11:58Z http://mathoverflow.net/feeds/question/122411 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/122411/almost-converses-to-the-am-gm-inequality Almost-converses to the AM-GM inequality Vincenzo 2013-02-20T15:28:54Z 2013-02-24T21:02:25Z <p>Let us consider the Arithmetic Mean -- Geometric Mean inequality for nonnegative real numbers: </p> <p>$$GM := (a_1 a_2 \ldots a_n)^{1/n} \le \frac{1}{n} \left( a_1 + a_2 + \ldots + a_n \right) =: AM.$$</p> <p>It is known that the converse inequality ($\ge$) holds if and only if all the $a_i$'s are the same. </p> <p>Therefore, we can expect that if the $a_i$'s are <em>almost</em> the same, then a converse inequality <em>almost</em> holds. For example, we may look for an inequality of the form $AM \le GM + f(\Delta,n)$ where $\Delta$ is the ratio between $\max_i a_i$ and $\min_i a_i$, but this is just one possibility. </p> <p>Are there any natural ways to formalize the above intuition? </p> http://mathoverflow.net/questions/122411/almost-converses-to-the-am-gm-inequality/122443#122443 Answer by Liviu Nicolaescu for Almost-converses to the AM-GM inequality Liviu Nicolaescu 2013-02-20T19:39:27Z 2013-02-24T18:12:47Z <p>Here is an old result of Siegel that is related to your question.</p> <p>Set</p> <p>$$s=s(a_1,\dotsc, a_n)=\frac{1}{n} (a_1+\cdots +a_n),$$</p> <p>$$p= p(a_1,\dotsc, a_n),$$</p> <p>$$\Delta= \Delta(a_1,\dotsc, a_n)=\prod_{i,j}(a_i-a_j)^2.$$</p> <p>The AM-GM inequality reads</p> <p>$$\frac{s^n}{p}\geq 1.$$</p> <p>Observe that $s$ is homogeneous of degree $1$, $p$ is homogeneous of degree $n$ and $\Delta$ is homogeneous of degree $n(n-1)$ in the variables $a_j$. In particular, the ration</p> <p>$$R= \frac{p^{n-1}}{\Delta}$$</p> <p>is homogeneous of degree $0$. Note that $\Delta=0$ when two of the numbers $a_j$ are equal. In particular, large $\Delta$ would mean that the numbers are "far from being equal". Equivalently the larger $\Delta$ is, the more "dispersed" are the numbers $a_j$.</p> <p>One can ask how dispersed can the numbers $a_j$ be given that $s$ and $p$ are fixed.</p> <p>In other words we ask to find</p> <p>$$\max \Delta(a_1,\dotsc, a_n)$$</p> <p>given that</p> <p>$$s(a_1,\dotsc, a_n)=s_0,\;\;p(a_1,\dotsc, a_n)=p_0.$$</p> <p>This constrained maximum exists and can be described <em>explicitly</em> as the discriminant of a certain Laguerre polynomial. I will denote it by $\Delta_\max(s_0,p_0)$. </p> <p>I will set</p> <p>$$\rho=\rho(s_0,p_0)= \frac{p_0^{n-1}}{\Delta_\max(s_0,p_0)}.$$</p> <p>Then there exists an <em>explicit</em> but <em>very complicated</em> strictly decreasing continuous function</p> <p>$$F_n: (0,\infty)\to (1,\infty)$$</p> <p>such that</p> <p>$$\lim_{t\to\infty} F_n(t)=1,$$</p> <p>$$\frac{s_0^n}{p_0}= F_n(\rho)= F_n\left( \frac{p_0^{n-1}}{\Delta_\max(s_0,p_0)}\right).$$</p> <p>A few things a bout the function $F_n$. It is described as a composition $Q_n\circ P_n^{-1}$, were </p> <p>$$Q_n: (0,\infty)\to (1,\infty)$$</p> <p>is a strictly decreasing <em>very explicit</em> rational function and </p> <p>$$P_n:(0,\infty)\to (0,\infty)$$</p> <p>is a <em>very explict</em> and strictly increasing polynomial such that $P_n(0)=0$. This implies the sharper inequality</p> <p>$$s(a_1, \dotsc, a_n)^n \geq F\left(\frac{p(a_1,\dotsc, a_n)^{n-1}}{\Delta(a_1,\dotsc, a_n)}\right)p(a_1,\dotsc, a_n),$$</p> <p>with equality iff</p> <p>$$\Delta(a_1,\dotsc,a_n)=\Delta_\max(s,p).$$</p> <p>For more details see Sec. 8.6 of the beautiful book <strong>Special Functions</strong> by G.E. Andrews, R. Askey, R. Roy.</p> http://mathoverflow.net/questions/122411/almost-converses-to-the-am-gm-inequality/122444#122444 Answer by Max Alekseyev for Almost-converses to the AM-GM inequality Max Alekseyev 2013-02-20T19:40:41Z 2013-02-21T15:44:12Z <p>Power mean inequality can give many bounds for the difference between AM and GM. Most simple is $$AM - GM \leq \max_i a_i - \min_i a_i.$$ Another bound is $$AM - GM \leq AM - HM = \frac{a_1+\dots+a_n}{n} - \frac{n}{1/a_1 + \dots + 1/a_n}$$ etc.</p> <p>See <a href="http://en.wikipedia.org/wiki/Generalized_mean#Generalized_mean_inequality" rel="nofollow">http://en.wikipedia.org/wiki/Generalized_mean#Generalized_mean_inequality</a></p> http://mathoverflow.net/questions/122411/almost-converses-to-the-am-gm-inequality/122446#122446 Answer by Mark Meckes for Almost-converses to the AM-GM inequality Mark Meckes 2013-02-20T19:53:08Z 2013-02-20T19:53:08Z <p>It's not precisely what you asked about, but <a href="http://link.springer.com/chapter/10.1007%2F978-3-540-36428-3_11?LI=true" rel="nofollow">this paper</a> by Gluskin and Milman shows that, for "most" sequences $a_1, \dotsc, a_n$, the AM-GM inequality can be reversed up to a multiplicative constant. The paper contains a number of observations which come closer to directly addressing your question.</p> http://mathoverflow.net/questions/122411/almost-converses-to-the-am-gm-inequality/122461#122461 Answer by Will Sawin for Almost-converses to the AM-GM inequality Will Sawin 2013-02-20T21:27:11Z 2013-02-20T21:27:11Z <p>The left and right sides are both continuous functions.</p> http://mathoverflow.net/questions/122411/almost-converses-to-the-am-gm-inequality/122828#122828 Answer by Felix Goldberg for Almost-converses to the AM-GM inequality Felix Goldberg 2013-02-24T21:02:25Z 2013-02-24T21:02:25Z <p>Proposition 1 in <a href="http://vuir.vu.edu.au/17286/" rel="nofollow">this paper</a> might be what you are looking for.</p>