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Triangle inequality squared...
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toh...@virgilio.it
Dec 29, 2008, 2:54:12 PM12/29/08
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I know by the triangle inequality that:
|a-b|=|a-c+c-b|<=|a-c|+|c-b|
However I have found in a book:
|a-b|^2 <= 2|a-c|^2 + 2|c-b|^2
How to prove this? I noticed only that could be false state that: |a-b|
^2 <= |a-c|^2 + |c-b|^2 (try with a=3, b=5, c=4). I have to take the
squares twice on the right side.
|a-b|=|a-c+c-b|<=|a-c|+|c-b|
However I have found in a book:
|a-b|^2 <= 2|a-c|^2 + 2|c-b|^2
How to prove this? I noticed only that could be false state that: |a-b|
^2 <= |a-c|^2 + |c-b|^2 (try with a=3, b=5, c=4). I have to take the
squares twice on the right side.
Any help appreciated so much.
Brian M. Scott
Dec 29, 2008, 3:17:06 PM12/29/08
to
On Mon, 29 Dec 2008 11:54:12 -0800 (PST),
<toh...@virgilio.it> wrote in
<news:9c5963e2-9645-4bbf...@a12g2000pro.googlegroups.com>
in alt.algebra.help:
<toh...@virgilio.it> wrote in
<news:9c5963e2-9645-4bbf...@a12g2000pro.googlegroups.com>
in alt.algebra.help:
0 <= (a + b - 2c)^2 = a^2 + 2ab + b^2 + 4c^2 - 4ac - 4bc, so
-2ab <= a^2 + b^2 + 4c^2 - 4ac - 4bc. Add a^2 + b^2 to both
sides:
(a + b)^2 = a^2 - 2ab + b^2 <=
2a^2 + 2b^2 + 4c^2 - 4ac - 4bc =
2a^2 - 4ac + 2c^2 + 2c^2 - 4bc + 2b^2 =
2(a - c)^2 + 2(c - b)^2.
Brian
toh...@virgilio.it
Dec 30, 2008, 8:45:26 AM12/30/08
to
Thank you a lot
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