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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm trying to put together an inductive proof of Cauchy's inequality for the complex case, $$ \left|\sum_{i=1}^na_ib_i\right|^2\leq\sum_{i=1}^n|a_i|^2\sum_{i=1}^n|b_i|^2. $$ The base case is easy, and at the inductive step I do as follows: $$ \begin{align*} \left|\sum_{i=1}^{n+1}a_ib_i\right|^2 =\left|\sum_{i=1}^n a_ib_i+a_{n+1}b_{n+1}\right|^2 &\leq\left(\left|\sum_{i=1}^n a_ib_i\right|+|a_{n+1}b_{n+1}|\right)^2\\ &= \left|\sum_{i=1}^n a_ib_i\right|^2+|a_{n+1}b_{n+1}|^2+2\left|\sum_{i=1}^na_ib_i\right||a_{n+1}b_{n+1}|\\ &\leq \sum_{i=1}^n|a_i|^2\sum_{i=1}^n |b_i|^2+|a_{n+1}b_{n+1}|^2+2|a_{n+1}b_{n+1}|\sum_{i=1}^n|a_i||b_i| \end{align*} $$ but I don't know what to do to conclude that the original sum is somehow less than or equal to $\sum_{i=1}^{n+1}|a_i|^2\sum_{i=1}^{n+1}|b_i|^2$. How can this be finished? Thanks.

share|cite|improve this question
up vote 2 down vote accepted

Let $s^2=\sum\limits_{i=1}^n|a_i|^2$, $t^2=\sum\limits_{i=1}^n |b_i|^2$, $a=|a_{n+1}|$ and $b=|b_{n+1}|$. The RHS of the last displayed inequality in the post is $R=s^2t^2+a^2b^2+2ab\sum\limits_{i=1}^n|a_i|\,|b_i|$.

First step: Cauchy-Schwarz inequality yields $\sum\limits_{i=1}^n|a_i|\,|b_i|\leqslant st$ hence $R\leqslant s^2t^2+a^2b^2+2abst$.

Second step: Since $2abst\leqslant a^2s^2+b^2t^2$, $R\leqslant s^2t^2+a^2b^2+a^2s^2+b^2t^2=(s^2+a^2)(t^2+b^2)$.

Conclusion: Since $s^2+a^2=\sum\limits_{i=1}^{n+1}|a_i|^2$ and $t^2+b^2=\sum\limits_{i=1}^{n+1} |b_i|^2$, this is the desired inequality.

Edit Note that $s^2t^2+a^2b^2+2abst=(st+ab)^2=\langle\sigma,\tau\rangle^2$ where $\sigma=(s,a)$, $\tau=(t,b)$ and $\langle\cdot,\cdot\rangle$ is the canonical scalar product. Hence an alternative to the second step is to use again Cauchy-Schwarz inequality, which yields $R\leqslant\langle\sigma,\tau\rangle^2\leqslant\langle\sigma,\sigma\rangle\cdot\langle\tau,\tau\rangle=(s^2+a^2)(t^2+b^2)$.

share|cite|improve this answer
Thank you Didier Piau. Why does $2abst\leq a^2s^2+b^2t^2$? – nicole jung Jan 18 '12 at 16:30
Do you recognize $(as)^2-2(as)(bt)+(bt)^2$? – Did Jan 18 '12 at 16:38
Oh right, that's $(as-bt)^2\geq 0$, thanks again. – nicole jung Jan 18 '12 at 16:40

Your Answer


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.