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

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- Jun 3rd 2006, 07:03 PMNichelle14Cauchy Sequence Inequality
I am to prove CS inequality by mathematical induction. Don't even know where to start. Any help?

- Jun 4th 2006, 07:03 AMmalaygoelQuote:

Originally Posted by**Nichelle14**

I don't care much about names so I can't figure it out. - Jun 4th 2006, 08:26 AMrgep
Perhaps it's Cauchy-

*Schwartz*? - Jun 4th 2006, 08:30 AMNichelle14
yes. sorry you are right it is Cauchey-Schwartz.

- Jun 4th 2006, 09:46 AMTD!
By induction? Induction on what then?

- Jun 4th 2006, 09:56 AMThePerfectHackerQuote:

Originally Posted by**TD!**

As I understand it she IS NOT talking about $\displaystyle \mathbb{R}^3$ but any dimesnion. - Jun 4th 2006, 11:13 AMrgep
The proof I usually use for Cauchy-Schwartz, in the form $\displaystyle \Vert u \Vert^2 \Vert v \Vert^2 \ge (u\cdot v)^2$, is to consider the quadratic $\displaystyle f(x) = \Vert u + xv \Vert^2 \ge 0$ 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 $\displaystyle (x^2+y^2)(u^2+v^2) = (xu+yv)^2 + (xv-yu)^2 \ge (xu+yv)^2$. The induction step would be something like $\displaystyle ((u_1^2+\cdots+u_{n-1}^2)+u_n^2)((v_1+\cdots+v_{n-1}^2)+v_n^2) \ge $ $\displaystyle \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 $ $\displaystyle \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. - Jun 4th 2006, 03:19 PMThePerfectHacker
I understand the problem as,

for any $\displaystyle n$,

$\displaystyle |x_1y_1+...+x_ny_n|\leq \sqrt{x_1^2+...+x_n^2}\cdot \sqrt{y_1^2+...+y_n^2}$

For any $\displaystyle x_i.y_i\in \mathbb{R}$