Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Is this part of my proof by induction correct ?

$\sum_{i=1}^{n}x_{i}y_{i}\leq \sqrt{\sum_{i=1}^{n}x_{i}^{2}\sum_{i=1}^{n}y_{i}^{2}}$

this is true when the true is that :

$\sum_{i=1}^{n}\left |x_{i}y_{i}\right |\leq \sqrt{\sum_{i=1}^{n}x_{i}^{2}\sum_{i=1}^{n}y_{i}^{2}}$

above inequality is true for $n=1$ and we assume that it's true for $n$.

For $n+1$ we get : $\sum_{i=1}^{n}\left |x_{i}y_{i}\right |+\left |x_{n+1}y_{n+1}\right |\leq \sqrt{\sum_{i=1}^{n}x_{i}^{2}\sum_{i=1}^{n}y_{i}^{2}+x_{n+1}^{2}\sum_{i=1}^{n}y_{i}^{2}+y_{n+1}^{2}\sum_{i=1}^{n}x_{i}^{2}+x_{n+1}^{2}y_{n+1}^{2}}$

using induction assumption we get :

$\sum_{i=1}^{n}\left |x_{i}y_{i}\right |+\left |x_{n+1}y_{n+1}\right |\leq \sqrt{\sum_{i=1}^{n}x_{i}^{2}\sum_{i=1}^{n}y_{i}^{2}}+\left |x_{n+1}y_{n+1}\right |\leq \sqrt{\sum_{i=1}^{n}x_{i}^{2}\sum_{i=1}^{n}y_{i}^{2}+x_{n+1}^{2}\sum_{i=1}^{n}y_{i}^{2}+y_{n+1}^{2}\sum_{i=1}^{n}x_{i}^{2}+x_{n+1}^{2}y_{n+1}^{2}}$

Is this correct ? Someone told me that I've used induction in wrong manner.

I'm adding link provided by http://math.stackexchange.com/users/18986/david-mitra http://ajmaa.org/RGMIA/papers/v12e/Cauchy-Schwarzinequality.pdf

share|improve this question
    
So many extra parentheses in these equation. Why write $(x_i)^2$ rather than $x_i^2$? –  Thomas Andrews May 7 '12 at 15:02
add comment

1 Answer

up vote 1 down vote accepted

The inductive step you made for the first inequality for the n+1 case while true is not necessary to show the 2nd key inequality to be true. Also, you haven't finished the proof, because the last inequality has not been shown to be true, i.e. reduced with n case or other true statements.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.