# Cauchy Sequence Inequality

• June 3rd 2006, 07:03 PM
Nichelle14
Cauchy Sequence Inequality
I am to prove CS inequality by mathematical induction. Don't even know where to start. Any help?
• June 4th 2006, 07:03 AM
malaygoel
Quote:

Originally Posted by Nichelle14
I am to prove CS inequality by mathematical induction. Don't even know where to start. Any help?

tell me the CS inequality.
I don't care much about names so I can't figure it out.
• June 4th 2006, 08:26 AM
rgep
Perhaps it's Cauchy-Schwartz?
• June 4th 2006, 08:30 AM
Nichelle14
yes. sorry you are right it is Cauchey-Schwartz.
• June 4th 2006, 09:46 AM
TD!
By induction? Induction on what then?
• June 4th 2006, 09:56 AM
ThePerfectHacker
Quote:

Originally Posted by TD!
By induction? Induction on what then?

The dimension on the vector space.

As I understand it she IS NOT talking about $\mathbb{R}^3$ but any dimesnion.
• June 4th 2006, 11:13 AM
rgep
The proof I usually use for Cauchy-Schwartz, in the form $\Vert u \Vert^2 \Vert v \Vert^2 \ge (u\cdot v)^2$, is to consider the quadratic $f(x) = \Vert u + xv \Vert^2 \ge 0$ and observe that the requirement ("b^2-4ac") that f has at most one real root is precisely C-S. This doesn't depend on dimension.

If I had to do it by induction I think I'ld start by observing that it is trivial for n=1. The case n=2 follows from $(x^2+y^2)(u^2+v^2) = (xu+yv)^2 + (xv-yu)^2 \ge (xu+yv)^2$. The induction step would be something like $((u_1^2+\cdots+u_{n-1}^2)+u_n^2)((v_1+\cdots+v_{n-1}^2)+v_n^2) \ge$ $\big\vert\sqrt{u_1^2+\cdots+u_{n-1}^2}\sqrt{v_1+\cdots+v_{n-1}^2}+u_nv_n \big\vert \ge$ $\big\vert(u_1,\ldots,n_{n-1})\cdot(v_1,\ldots,v_{n-1})+u_nv_n\big\vert$, thus using the cases 2 and n-1.
• June 4th 2006, 03:19 PM
ThePerfectHacker
I understand the problem as,
for any $n$,
$|x_1y_1+...+x_ny_n|\leq \sqrt{x_1^2+...+x_n^2}\cdot \sqrt{y_1^2+...+y_n^2}$
For any $x_i.y_i\in \mathbb{R}$