Discrete Mathematics and Its Applications, Seventh Edition

Published by McGraw-Hill Education
ISBN 10: 0073383090
ISBN 13: 978-0-07338-309-5

Chapter 1 - Section 1.8 - Proof Methods and Strategy - Exercises - Page 108: 23

Answer

The geometric mean is greater or equal to the harmonic mean

Work Step by Step

Let $x$ and $y$ be two real numbers, $x>0,y>0$. The geometric mean is: $M_g=\sqrt{xy}$, while the harmonic mean is: $M_h=\dfrac{2}{\frac{1}{x}+\frac{1}{y}}=\dfrac{2xy}{x+y}$ We have: $(x-y)^2\geq 0$ $x^2-2xy+y^2\geq 0$ $x^2-2xy+y^2+4xy\geq 0+4xy$ $x^2+2xy+y^2\geq 4xy$ $(x+y)^2\geq 4xy$ $xy(x+y)^2\geq xy(4xy)$ $xy(x+y)^2\geq 4x^2y^2$ $\dfrac{xy(x+y)^2}{(x+y)^2}\geq \dfrac{4x^2y^2}{(x+y)^2}$ $xy\geq \left(\dfrac{2xy}{x+y}\right)^2$ $\sqrt{xy}\geq \dfrac{2xy}{x+y}$ $M_g\geq M_h$ Therefore the geometric mean is greater or equal to the harmonic mean.
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