By Zdravko Cvetkovski
This paintings is set inequalities which play a massive function in mathematical Olympiads. It comprises one hundred seventy five solved difficulties within the type of workouts and, additionally, 310 solved difficulties. The publication additionally covers the theoretical history of an important theorems and strategies required for fixing inequalities. it really is written for all center and high-school scholars, in addition to for graduate and undergraduate scholars. college lecturers and running shoes for mathematical competitions also will achieve make the most of this book.
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Extra info for Inequalities: Theorems, Techniques and Selected Problems
Sample text
We have (a + b + c)(a 2 + b2 + c2 ) ≤ 3(a 3 + b3 + c3 ) ⇔ a 3 + b 3 + c 3 a 2 + b2 + c 2 ≥ . a+b+c 3 Similarly we get a 3 + b3 + d 3 a 2 + b2 + d 2 ≥ , a+b+d 3 a 3 + c3 + d 3 a 2 + c 2 + d 2 ≥ , a+c+d 3 b3 + c3 + d 3 b2 + c 2 + d 2 ≥ . b+c+d 3 After adding these inequalities we get the required inequality. 19 Let a1 , a2 , . . , an ∈ R+ such that a1 + a2 + · · · + an = 1. Prove the inequality a2 an n a1 . + + ··· + ≥ 2 − a1 2 − a2 2 − an 2n − 1 Solution Without loss of generality we may assume that a1 ≥ a2 ≥ · · · ≥ an .
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. e. 1) holds for n = k + 1, and we are done. Z. 1 (Bernoulli’s inequality) Let n ∈ N and x > −1. Then (1 + x)n ≥ 1 + nx. 1, for x1 = x2 = · · · = xn = x, we obtain the required result. 1 We’ll say that the function f (x1 , x2 , . . , xn ) is homogenous with coefficient of homogeneity k, if for arbitrary t ∈ R, t = 1, we have f (tx1 , tx2 , . .
22) 9 12s 4s · = . 18 Let a, b, c, d ∈ R+ . Prove the inequality a 3 + b 3 + c3 a 3 + b 3 + d 3 a 3 + c3 + d 3 b3 + c3 + d 3 + + + a+b+c a+b+d a+c+d b+c+d ≥ a 2 + b2 + c 2 + d 2 . Solution Without loss of generality we may assume that a ≥ b ≥ c ≥ d. Then clearly a 2 ≥ b2 ≥ c2 ≥ d 2 . e. we have (a + b + c)(a 2 + b2 + c2 ) ≤ 3(a 3 + b3 + c3 ) ⇔ a 3 + b 3 + c 3 a 2 + b2 + c 2 ≥ . a+b+c 3 Similarly we get a 3 + b3 + d 3 a 2 + b2 + d 2 ≥ , a+b+d 3 a 3 + c3 + d 3 a 2 + c 2 + d 2 ≥ , a+c+d 3 b3 + c3 + d 3 b2 + c 2 + d 2 ≥ .