By Zdravko Cvetkovski

This paintings is set inequalities which play a massive function in mathematical Olympiads. It includes one hundred seventy five solved difficulties within the kind of routines and, moreover, 310 solved difficulties. The booklet additionally covers the theoretical history of an important theorems and methods required for fixing inequalities. it truly is written for all heart and high-school scholars, in addition to for graduate and undergraduate scholars. tuition academics and running shoes for mathematical competitions also will achieve make the most of this book.

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1 Points of Incidence in Applications of the AM–GM Inequality In this subsection we will consider characteristic examples in which we can use incorrectly the inequality AM ≥ GM. Namely, a possible major route for the proper use of this inequality (the means inequalities) will be the fact that equality in these inequalities is achieved when all variables are equal. These points at which equality (all their coordinates are equal) of a given inequality is satisfied are called points of incidence. It is also important to note that symmetrical expressions achieve a minimum or maximum at a point of incidence.

Then we have (1 + x1 )(1 + x2 ) · · · (1 + xn ) ≥ 1 + x1 + x2 + · · · + xn . 1) Proof We’ll prove the given inequality by induction. For n = 1 we have 1 + x1 ≥ 1 + x1 . Suppose that for n = k, and arbitrary real numbers xi > −1, i = 1, 2, . . e. (1 + x1 )(1 + x2 ) · · · (1 + xk ) ≥ 1 + x1 + x2 + · · · + xk . 2) Let n = k + 1, and xi > −1, i = 1, 2, . . , k + 1, be arbitrary real numbers with the same signs. Then, since x1 , x2 , . . , xk+1 have the same signs, we have (x1 + x2 + · · · + xk )xk+1 ≥ 0.

Iff x1 = x2 = · · · = xk+1 = 1. 3) is proved. e. n n n 1 2 1 2 a1 = a2 = · · · = an . 1), and we are done. e. √ n a1 a2 · · · an ≥ n 1 a1 + + ··· + 1 a2 1 an . e. e. e. a1 = a2 = · · · = an . a12 + a22 + · · · + an2 a1 + a2 + · · · + an ≥ . n n We’ll use the Cauchy–Schwarz inequality for the sequences (a1 , a2 , . . , an ) and (1, 1, . . , 1). So we have (a12 + a22 + · · · + an2 )(12 + 12 + · · · + 12 ) ≥ (a1 + a2 + · · · + an )2 ⇔ n(a12 + a22 + · · · + an2 ) ≥ (a1 + a2 + · · · + an )2 ⇔ a12 + a22 + · · · + an2 ≥ n ⇔ 2 a1 + a2 + · · · + an n a12 + a22 + · · · + an2 a1 + a2 + · · · + an ≥ .

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