# Inequalities : a Mathematical Olympiad approach by Radmila Bulajich Manfrino By Radmila Bulajich Manfrino

This booklet is meant for the Mathematical Olympiad scholars who desire to organize for the learn of inequalities, a subject now of common use at a variety of degrees of mathematical competitions. during this quantity we current either vintage inequalities and the extra worthy inequalities for confronting and fixing optimization difficulties. a huge a part of this publication offers with geometric inequalities and this truth makes a major distinction with recognize to many of the books that take care of this subject within the mathematical olympiad. The booklet has been equipped in 4 chapters that have each one of them a unique personality. bankruptcy 1 is devoted to provide uncomplicated inequalities. such a lot of them are numerical inequalities as a rule missing any geometric which means. despite the fact that, the place it's attainable to supply a geometrical interpretation, we comprise it as we pass alongside. We emphasize the significance of a few of those inequalities, comparable to the inequality among the mathematics suggest and the geometric suggest, the Cauchy-Schwarz inequality, the rearrangementinequality, the Jensen inequality, the Muirhead theorem, between others. For most of these, in addition to giving the evidence, we current a number of examples that express tips to use them in mathematical olympiad difficulties. We additionally emphasize how the substitution technique is used to infer numerous inequalities.

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Extra info for Inequalities : a Mathematical Olympiad approach

Example text

Xn , y1 , y2 , . . , yn , z1 , z2 , . . , zn be positive real numbers. (i) If a, b, c are positive real numbers such that 1 c n (xi yi ) xi i=1 n xi yi zi ≤ i=1 yi 1 a + 1 b n xi a i=1 1 b n b . i=1 (ii) If a, b, c are positive real numbers such that n = 1c , then a i=1 1 a 1 b + 1 a n ≤ c 1 a 1 b + 1 c 1 c n yi b i=1 = 1, then zi c . 89. 90. Let a, b, c be non-negative real numbers. Prove that √ 3 (i) a2 + b2 + c2 + 3 a2 b2 c2 ≥ 2(ab + bc + ca), (ii) a2 + b2 + c2 + 2abc + 1 ≥ 2(ab + bc + ca).

Making an adequate substitution we can, for instance, change the diﬃcult terms of the inequality a little, we can simplify expressions or we can reduce terms. In this section we give some ideas of what can be done with this strategy. As always, the best way to do that is through some examples. One useful suggestion for problems that contain in the hypothesis an extra condition, is to use that condition to simplify the problem. In the next example we apply this technique to eliminate the denominators in order to make the problem easier to solve.

F (x1 , . . , xn ) the sum of the n! terms obtained from evaluating F in all possible permutations of (x1 , . . , xn ). We will consider only the particular case F (x1 , . . , xn ) = xa1 1 xa2 2 · · · xann We write [a] = [a1 , a2 , . . , an ] = ables x, y, z > 0 we have that 1 n! [1, 1] = xy, [1, 1, 1] = xyz, [2, 1, 0] = with xi > 0, ai ≥ 0. xa1 1 xa2 2 · · · xann . For instance, for the vari1 2 [x (y + z) + y 2 (x + z) + z 2 (x + y)]. 3! 44 Numerical Inequalities It is clear that [a] is invariant under any permutation of the (a1 , a2 , .