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 a_r ,b_r ,c_r where r=1,2,3 are non-negative real numbers and

\sum_{r = 1}^{3} (a_{r} + b_{r} + c_{r}) = 3 L

show that $V \leq L^{3}$
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 a_i .

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 \times b) . c

(standard formula)

$| r \times s | = | r | | s | \sin θ < | r | | s |$ 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.