By J. Michael Steele

Michael Steele describes the basic subject matters in mathematical inequalities and their makes use of. utilizing the Cauchy-Schwarz inequality as a consultant, Steele offers a desirable selection of difficulties relating to inequalities and coaches readers via strategies, in a method akin to George Polya, through educating uncomplicated recommendations and sprucing challenge fixing talents even as. Undergraduate and starting graduate scholars in arithmetic, theoretical desktop technology, information, engineering, and economics will locate the booklet applicable for self-study.

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**Extra info for The Cauchy-Schwarz Master Class: An Introduction to the Art of Mathematical Inequalities (Maa Problem Books Series.)**

**Example text**

A box in Rn has 2n corners, and each of those corners is incident to n edges of the box. If we let the lengths of those edges be a1 , a2 , . . , an , then the same isoperimetric intuition that we have used for squares and cubes suggests that the n-cube with edge length S/n will have the largest volume among all boxes for which a1 + a2 + · · · + an = S. The next challenge problem oﬀers an invitation to ﬁnd an honest proof of this intuitive claim. It also recasts this geometric conjecture in the more common analytic language of arithmetic and geometric means.

Bn ), but one can think of other measures of proportionality that are just as reasonable. For example, if we restrict our attention to vectors of positive terms, then one might equally well use the self-normalized sum 1 R= 2 n n i=1 j=1 (ai bj − aj bi )2 . A. Milne: n n (aj + bj ) j=1 j=1 aj bj (aj + bj ) n ≤ n aj j=1 bj . 17) if and only if the vectors (a1 , a2 , . . , an ) and (b1 , b2 , . . , bn ) are proportional. 17) was introduced by Milne in 1925 to help explain the biases inherent in certain measurements of stellar radiation.

The natural plan is to take an n < 2k and to look for some way to use the n numbers a1 , a2 , . . , an to deﬁne a longer sequence α1 , α2 , . . 5). The discovery of an eﬀective choice for the values of the sequence {αi } may call for some exploration, but one is not likely to need too long to hit on the idea of setting αi = ai for 1 ≤ i ≤ n and setting a1 + a2 + · · · + an ≡A for n < i ≤ 2k ; αi = n in other words, we simply pad the original sequence {ai : 1 ≤ i ≤ n} with enough copies of the average A to give us a sequence {αi : 1 ≤ i ≤ 2k } that has length equal to 2k .