Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Appendices - Section A.1 - Real Numbers and the Real Line - Exercises A.1 - Page AP-6: 24


See below.

Work Step by Step

Step (1) applies the fact that $(-x)^{2}=(-1\cdot x)^{2}=(-1)^{2}x^{2}=1\cdot x^{2}=x^{2}$ (the square of a number is nonnegative, whether the number is negative or not, and, the squares of a number and its opposite number are equal in magnitude.) So, given that $|a+b|$ is always nonnegative and $(a+b) $ could be either negative or nonnegative, their squares will be equal in any case. Step (2) applies the fact that the product $ab$ could be negative or nonnegative. If it is negative, it is less than $|a|\cdot|b|$, a product of nonnegative numbers. If it it is nonnegative, it equals $|a|\cdot|b|$, a product of nonnegative numbers. In either case $ab\leq|a|\cdot|b|$, so adding $2ab$ to $a^{2}+b^{2}$, we get less than or equal to the case where we add $2|a|\cdot|b|$ to $a^{2}+b^{2}$ Step (3) applies the same principle as step 1, $|a|^{2}=a^{2}$ and $|b^{2}|=b^{2}$. Step(4) Taking the square root of both sides, with both sides being nonnegative, we apply the inequality $''$LHS$\leq$RHS$''$ , achieved in step (2), because $x_{1}\leq x_{2}\Rightarrow\sqrt{x_{1}}\leq\sqrt{x_{2}}$, for positive numbers $x_{1}$ and $x_{2}.$ (The square root function preserves order.)
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