Parallelepiped inequality
By Arun Iyer on Sunday, May 12, 2002 - 04:38 am:

can anyone help me here....

let V be the volume of the paralellepiped formed by the vectors
a=a1 i+a2 j+a3 k
b=b1 i+b2 j+b3 k
c=c1 i+c2 j+c3 k

If ar ,br ,cr where r=1,2,3 are non-negative real numbers and


3
å
r=1 
 (ar+br+cr)=3L

show that V £ L3
love arun

By David Loeffler on Sunday, May 12, 2002 - 08:38 am:

Arun,

For obvious geometrical reasons we have

V2 < = (a1 2 +a2 2 +a3 2 )(b1 2 +b2 2 +b3 2 )(c1 2 +c2 2 +c3 2 )

Now we note that (a1 2 + a2 2 +a3 2 ) < = (a1 +a2 +a3 )2 for non-negative reals ai .

So
V < = (a1 +a2 +a3 )(b1 +b2 +b3 )(c1 +c2 +c3 )

and your result is now immediate by AM-GM.

David

By Arun Iyer on Sunday, May 12, 2002 - 11:26 am:

David,
sorry I could not quite understand your first statement...can you elaborate on it please???

love arun

By David Loeffler on Sunday, May 12, 2002 - 01:09 pm:
V = (a ×b).c (standard formula)

|r ×s| = |r| |s| sinq < |r| |s| and |r.s| < |r| |s| similarly. Put these two together + result follows.

David

By Arun Iyer on Sunday, May 12, 2002 - 05:17 pm:

gotcha!!!
Thanks David...
I understood that line of proof....

Though, I wonder why you say "for obvious geometrical reasons..." is there any geometrical interpretation for that result....

love arun

By David Loeffler on Sunday, May 12, 2002 - 06:38 pm:

These were my obvious reasons; this is how I tend to think of triple scalar products, as volumes of parallelepipeds.