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】![](/moin_static193/solenoid/img/up.png) : 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】![](/moin_static193/solenoid/img/up.png) : 没头绪,哪位给个hint? : thanks for sharing!
|
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】![](/moin_static193/solenoid/img/up.png) : 先平方.
|
D**u 发帖数: 204 | 5 A lot of terms are canceled out.
【在 B********e 的大作中提到】![](/moin_static193/solenoid/img/up.png) : 哦,i give up : 一见一大堆平方我就有畏难情绪哈
|
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】![](/moin_static193/solenoid/img/up.png) : A lot of terms are canceled out.
|
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】![](/moin_static193/solenoid/img/up.png) : : 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】![](/moin_static193/solenoid/img/up.png) : 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).
|
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】![](/moin_static193/solenoid/img/up.png) : 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).
|
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】![](/moin_static193/solenoid/img/up.png) : 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
|
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】![](/moin_static193/solenoid/img/up.png) : (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:
|