# Inductive proof of Cauchy's inequality for complex numbers?

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.

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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 $R\leqslant\langle\sigma,\tau\rangle^2\leqslant\langle\sigma,\sigma\rangle\cdot\langle\tau,\tau\rangle=(s^2+a^2)(t^2+b^2)$.

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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