Prove the Arithmetic-Geometric Mean Inequality If a_1 a_2,..., a_n are nonnegative numbers, then their arithmetic mean is frac{a_1 + a_2 + ... + a_n}{

shadsiei 2020-11-29 Answered
Prove the Arithmetic-Geometric Mean Inequality If a1a2,,an are nonnegative numbers, then their arithmetic mean is a1+a2++ann, and their geometric . The avthmetic-geometic mean is n{a1,a2,an}. The arithmetic-geometric mean equality states that the geometric mean is always less than or equal to the arithmetic mean. In this problem we prove this in the case of two numbers andy. (a) If x and y are nonnegative and xy,thenx2y2. [Hint: First use Rule 3 of Inequalities to show that x2xyandxyy2. ] (b) Prove the arithmetic-geometric mean inequality xyx+y2
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Alix Ortiz
Answered 2020-11-30 Author has 109 answers

We have to prove inequalities according the question. a) If x and y are non-negative and xy ,then we want to prove x2y2. Here we have given, xy Since, x is non-negative then, x.xx.yx2xy ...(i)
Since, y is non-negative then, x.yy.yxyy2 ..(ii)
So,from equation (i) and (ii) we get x2y2 b) For any two non-negative numbers x and y we have (xy)20 Then, (xy)20x2+y22xy0x2+y2+2xy4xy[add 4xy onh sides](x+y)24xy(x+y)4xy(x+y)2xy(x+y)2xy
So, for any two non-negative number we have, xyx+y2

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Jeffrey Jordon
Answered 2022-01-15 Author has 2064 answers

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