20071010, 05:38  #1 
"Bo Chen"
Oct 2005
Wuhan,China
10101010_{2} Posts 
Ask an Advanced Algebra question.
a,b,c is Euclid space's three vectors.
Then: abc<=bca+cab It takes me a long time, but I still dont know how to prove it. I dont know I can ask this type question here,though this is a forum to discuss about mersenne prime. 
20071010, 12:49  #2  
Nov 2003
7460_{10} Posts 
Quote:
Think "triangle inequality". Hint 2: Divide by abc 

20071011, 04:47  #3 
"Bo Chen"
Oct 2005
Wuhan,China
2·5·17 Posts 
I know the following result
xy <= xy <= x+y (ab) = (ac)+(cb) (ab,c) = (ac,b)+(cb,a) but I still can't prove the conclusion. I can't understand the meaning of "Hint 2: Divide by abc". abc<=bca+cab <=> ab/(ab)<=bc/(bc)+ca/(ca) There seems no apparent connection among (ab)/(ab), (bc)/(bc) and (ca)/(ca). 
20071011, 07:08  #4  
Jan 2006
JHB, South Africa
157 Posts 
Quote:
(ab)/(ab) = (a)/(ab)  (b)/(ab) doing the same for the RHS. Then simplifying the result. Regards Patrick 

20071013, 11:47  #5 
"Lucan"
Dec 2006
England
1100101001010_{2} Posts 
Call me lazy, but I haven't nailed this one yet.

20071014, 09:46  #6 
"Nancy"
Aug 2002
Alexandria
9A3_{16} Posts 
If c has the smallest norm of the three, we can write
abc ≤ bcc + cac, assume c != 0 (or the inequality trivially holds) and divide by c ab ≤ bc + ca ⇐ ab ≤ (bc) + (ca) ⇔ ab ≤ ba So this case works. If any vector is 0 or any two vectors are identical, it also works. The rest eludes me. Alex 
20071015, 09:00  #7 
"Lucan"
Dec 2006
England
2×3×13×83 Posts 
A proof in one dimension might be a good start.
A,B,C all lie on the x axis. 
20071015, 10:19  #8 
Jan 2006
JHB, South Africa
157 Posts 
Dividing by abc
Code:
LHS: ab/(ab) = a/(ab)  b/(ab) = 1/b  1/a  RHS bc/(bc)+ca/(ca) = b/(bc)  c/(bc) + c/(ca)  a/(ca) thus 1/b  1/a  <= 1/c 1/b + 1/a  1/c 1/b  1/a  <= 1/b + 1/a  1/b  1/a  <= 1/a  1/b  Last fiddled with by Patrick123 on 20071015 at 10:21 
20071016, 00:49  #9  
"Bo Chen"
Oct 2005
Wuhan,China
170_{10} Posts 
Quote:
for example, a=(1,0),b=(0,1) then ab=(1,1) ab/(ab)=sqrt(2), but 1/b  1/a  = 11 = 0 

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