4
$\begingroup$

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 https://math.stackexchange.com/users/18986/david-mitra http://ajmaa.org/RGMIA/papers/v12e/Cauchy-Schwarzinequality.pdf

$\endgroup$
  • $\begingroup$ So many extra parentheses in these equation. Why write $(x_i)^2$ rather than $x_i^2$? $\endgroup$ – Thomas Andrews May 7 '12 at 15:02
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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