D**u
发帖数: 204 | 1 This is problem I saw many years ago, a fairly good one.
Let X,Y,Z be vectors in R^3. Prove that:
|X| + |Y| + |Z| + |X+Y+Z| >= |Y+Z| + |X+Z| + |X+Y|. |
B********e
发帖数: 10014 | 2 没头绪，哪位给个hint？
thanks for sharing!
【在 D**u 的大作中提到】
: This is problem I saw many years ago, a fairly good one.
: Let X,Y,Z be vectors in R^3. Prove that:
: |X| + |Y| + |Z| + |X+Y+Z| >= |Y+Z| + |X+Z| + |X+Y|.
|
D**u
发帖数: 204 | 3 先平方.
【在 B********e 的大作中提到】
: 没头绪，哪位给个hint？
: thanks for sharing!
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B********e
发帖数: 10014 | 4 哦，i give up
一见一大堆平方我就有畏难情绪哈
【在 D**u 的大作中提到】
: 先平方.
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D**u
发帖数: 204 | 5 A lot of terms are canceled out.
【在 B********e 的大作中提到】
: 哦，i give up
: 一见一大堆平方我就有畏难情绪哈
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r******o
发帖数: 122 | 6
请说下一步..
兄台数学太好，讲话不必落实，但仍希望能照顾一下平实的小民。
【在 D**u 的大作中提到】
: A lot of terms are canceled out.
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D**u
发帖数: 204 | 7 Take square of both sides of the original equation. Some terms are canceled
out, and after some re-arranging of the remaining terms, the inequality
becomes:
(|X|*|X+Y+Z|+|Y|*|Z|)+(|Y|*|X+Y+Z|+|X|*|Z|)+(|Z|*|X+Y+Z|+|X|*|Y|)
>= |X+Y|*|X+Z|+|Y+X|*|Y+Z|+|Z+X|*|Z+Y| (1)
Now if we can prove
|X|*|X+Y+Z|+|Y|*|Z| >= |X+Y|*|X+Z| (2)
|Y|*|X+Y+Z|+|X|*|Z| >= |Y+X|*|Y+Z| (3)
|Z|*|X+Y+Z|+|X|*|Y| >= |Z+X|*|Z+Y| (4)
then we are done by adding up (2),(3)and(4).
By symmetry, (2),(3) a
【在 r******o 的大作中提到】
:
: 请说下一步..
: 兄台数学太好，讲话不必落实，但仍希望能照顾一下平实的小民。
|
B********e
发帖数: 10014 | 8 试了下，但搞得4，5个大西格玛的term一堆，实在懒得搞哇
有什么trick吗？或者怎么看比较直观？
哎，太愚昧，太愚昧啦
canceled
【在 D**u 的大作中提到】
: Take square of both sides of the original equation. Some terms are canceled
: out, and after some re-arranging of the remaining terms, the inequality
: becomes:
: (|X|*|X+Y+Z|+|Y|*|Z|)+(|Y|*|X+Y+Z|+|X|*|Z|)+(|Z|*|X+Y+Z|+|X|*|Y|)
: >= |X+Y|*|X+Z|+|Y+X|*|Y+Z|+|Z+X|*|Z+Y| (1)
: Now if we can prove
: |X|*|X+Y+Z|+|Y|*|Z| >= |X+Y|*|X+Z| (2)
: |Y|*|X+Y+Z|+|X|*|Z| >= |Y+X|*|Y+Z| (3)
: |Z|*|X+Y+Z|+|X|*|Y| >= |Z+X|*|Z+Y| (4)
: then we are done by adding up (2),(3)and(4).
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B********e
发帖数: 10014 | 9 一双双明亮的眼睛盯着，却见死不救哈
准备烧马尾巴呼唤双修侠啦
canceled
【在 D**u 的大作中提到】
: Take square of both sides of the original equation. Some terms are canceled
: out, and after some re-arranging of the remaining terms, the inequality
: becomes:
: (|X|*|X+Y+Z|+|Y|*|Z|)+(|Y|*|X+Y+Z|+|X|*|Z|)+(|Z|*|X+Y+Z|+|X|*|Y|)
: >= |X+Y|*|X+Z|+|Y+X|*|Y+Z|+|Z+X|*|Z+Y| (1)
: Now if we can prove
: |X|*|X+Y+Z|+|Y|*|Z| >= |X+Y|*|X+Z| (2)
: |Y|*|X+Y+Z|+|X|*|Z| >= |Y+X|*|Y+Z| (3)
: |Z|*|X+Y+Z|+|X|*|Y| >= |Z+X|*|Z+Y| (4)
: then we are done by adding up (2),(3)and(4).
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D**u
发帖数: 204 | 10 (claim: a friend of mine gave the proof of this question, which is different from the "official" proof of the original question we saw in publication).
Now let's prove
|Y|*|X+Y+Z|+|X|*|Z| >= |X+Y|*|Y+Z| (3)
Proof:
Find a 四面体 ABCD such that
X = B-A, Y = C-B, Z=D-C.
Then (3) can be approved by directly applying 托勒密不等式
which says that "四边形的任两组对边乘积之和不小于另外一组对边的乘积，取等号当且
仅当共圆或共线。(四点可不限于同一平面)"
See the following links about 托勒密不等式:
http://zhidao.baidu.com/question/20356987.html
http://zh.wikipedia.org/wi
【在 B********e 的大作中提到】
: 试了下，但搞得4，5个大西格玛的term一堆，实在懒得搞哇
: 有什么trick吗？或者怎么看比较直观？
: 哎，太愚昧，太愚昧啦
:
: canceled
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B********e
发帖数: 10014 | 11 great! man, i appreciate it!
different from the "official" proof of the original question we saw in
publication).
号当且
【在 D**u 的大作中提到】
: (claim: a friend of mine gave the proof of this question, which is different from the "official" proof of the original question we saw in publication).
: Now let's prove
: |Y|*|X+Y+Z|+|X|*|Z| >= |X+Y|*|Y+Z| (3)
: Proof:
: Find a 四面体 ABCD such that
: X = B-A, Y = C-B, Z=D-C.
: Then (3) can be approved by directly applying 托勒密不等式
: which says that "四边形的任两组对边乘积之和不小于另外一组对边的乘积，取等号当且
: 仅当共圆或共线。(四点可不限于同一平面)"
: See the following links about 托勒密不等式:
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