Prove |x+y| ≥ |x| - |y|. It gives a hint to turn x=x+y-y and to use |a + b| ≤ |a| + |b| (triangle inequality) with the fact |-y| = |y|, but I'm not sure where to start.
Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Not the answer you are looking for? Search for more explanations.
since its -2 * magnitude of x and y then
its always a negative number or 0 when x and y = 0
so this equality holds true
2YX can only be greater or equal to -2}y}}x}
\(|x| = |(x-y)+y|\le |(x-y)| + |y|\).
therefore, removing |y| on both sides: \(|x|-|y|\le |x-y| \). Then you just have to be able to explain why you can replace \(y\) by \(-y\). and you're done.
Would this work?
Let a = x + y, b = -y
i) |a + b| ≤ |a| + |b| (triangle inequality)
ii) |(x+y) - y| ≤ |x+y| + |-y| (substitution)
iii) |x| - |-y| ≤ |x+y| (subtract |-y| from both sides)
iv) |x| - |y| ≤ |x+y| (|-y| = |y| definition of absolute value)
exactly how you should do it. well done.
Thanks to both of you. What reemii wrote gave me a flash of insight. However, the way the hint is written in the book was confusing as I was literally substituting a in the triangle inequality with a = x + y - y which was confusing me for a long time.