From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik) Subject: Re: (1+x)^n >= 1+nx Date: 28 May 2000 16:24:01 -0400 Newsgroups: sci.math Summary: Arithmetic-Geometric Mean Inequalities (AGMI) In article , Carl Johan Ragnarsson wrote: :>By a medium-clever re-arrangement, you can use Bernoulli inequality to :>prove the Arithmetic-Geometric Mean Inequality (AGM), one variable at a :>time. The details are too messy for a usenet post. But AGM is useful in :>finding extrema of some functions without calculus. : :I never saw this proof, nor do I find it obvious how to use the :inequality. :Any brief sketch, or reference? It is not in Beckenbach-Bellman, I think. And I might not be the first to come up with it. If so, you are witnessing the first publication for a wider audience (other than a high-school coaching session). :-)= And even a brief sketch is messy. You asked for it - and typos might have occurred during transcription from EXP to ASCII: Arithmetic-Geometric Mean Inequalities (AGMI) Z.V. Kovarik, 1997 or before. Two inductive proofs, one variable at a time. Let a(1),...,a(N) be positive numbers. For n=1,...,N define A(n)=(a(1)+...+a(n))/n and G(n)=(a(1)*...*a(n))^(1/n) (1) First refinement of AGMI: For n=1,...,N-1, (A(n+1)/(G(n+1))^(n+1) >= (A(n)/G(n))^n with equality if and only if a(n+1)=A(n). (2) Second refinement of AGMI: For n=1,...,N-1, (n+1)*(A(n+1)-G(n+1)) >= n*(A(n)-G(n)) with equality if and only if a(n+1)=G(n). (It is obvious that A(1)=G(1)=a(1), and each refinement implies A(N) >= G(N) with equality if and only if all a(n)=a(1).) Recall Bernoulli Inequality: If t>-1, t not 0, and if n is an integer, n>=2 then (1+t)^n > 1+n*t. Proof of (1): Algebra part: Denote r(n)=(A(n)/G(n))^n (we want to compare r(n+1) with r(n)), then one can check that r(n+1)/r(n) = (A(n)/a(n+1))*(1+(a(n+1)-A(n))/((n+1)*A(n)))^(n+1) Inequality part: Except when a(n+1)=A(n), the second factor is greater than 1+(a(n+1)-A(n))/A(n) = a(n+1)/A(n) and finally r(n+1)/r(n) > (A(n)/a(n+1)) * a(n+1)/A(n) = 1. So, r(n+1) > r(n) unless a(n+1)=A(n), as claimed. Proof of (2): Algebra part: Denote D(n)=n*(A(n)-G(n)) ; we want to compare D(n+1) with D(n). The difference D(n+1)-D(n) simplifies to D(n+1)-D(n)=a(n+1)-(n+1)*G(n+1)+n*G(n) Introduce more notation: let a(n+1)=b^(n+1) and G(n)=H^(n+1) so that conveniently (check it!) G(n+1)=b*H^n and we have D(n+1)-D(n)=... =H^(n+1) * (1+(b-H)/H)^(n+1) - (n+1)*b*H^n + n*H^(n+1) (*) Inequality part: Unless b=H, the first term is greater than H^(n+1) * (1+(n+1)*(b-H)/H) which exactly cancels the rest of the expression in (*). The condition b=H is equivalent to a(n+1)=G(n); unless it takes place, we have D(n+1) > D(n), as claimed. ------------------------- Remark about "how on earth did he discover the inequalities in the first place?": Solve the Calculus problem: given positive a(1),...,a(n), define for t>0 the function f(t) = ((a(1)+...+a(n)+t)/(n+1) / (a(1)*...*a(n)*t)^(1/(n+1)) - it is, let a(n+1) "float around", and try to find the minimum of f. Since AGMI was proved many times before, the minimum is supposed to be no less than 1. Locating and evaluating the minimum leads to the first refinement. The second refinement is obtained in a similar way. Have fun, ZVK(Slavek). ============================================================================== From: ikastan@uranus.uucp (Ilias Kastanas) Subject: Re: (1+x)^n >= 1+nx Date: 28 May 2000 21:08:14 GMT Newsgroups: sci.math In article , Carl Johan Ragnarsson wrote: @>By a medium-clever re-arrangement, you can use Bernoulli inequality to @>prove the Arithmetic-Geometric Mean Inequality (AGM), one variable at a @>time. The details are too messy for a usenet post. But AGM is useful in @>finding extrema of some functions without calculus. @ @I never saw this proof, nor do I find it obvious how to use the inequality. @Any brief sketch, or reference? Chrystal's "Algebra" (1889). Ilias