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Deduce the Cauchy-Schwarz Inequality from the case m = 1 of Bessel’s Inequality:
the sum of $$\sum_{i=1}^{m}|(v,u_i)|^2 \leq ||v||^2. $$

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I think you mean $m=0$, or else start the sum at $i=1$. – Siméon Apr 12 '13 at 9:21
I believe $u_i$ belongs to the orthonormal basis which means $||u_i||=1$. So, just apply the Cauchy-Schwartz inequality and you will get the answer. – Mhenni Benghorbal Apr 12 '13 at 11:41

Let $a,b$ be two vectors.

If $b = 0$, the inequality $|(a,b)| \leq \|a\|\|b\|$ is trivially true.

If $b \neq 0$ then $u = \|b\|^{-1}b$ has norm $1$, hence by Bessels's inequality with $m=0$ we have $|(a,u)|^2 \leq \|a\|^2$. Since $(a,b) = (a,\|b\|u) = \|b\| (a,u)$, this yields $|(a,b)| \leq \|a\| \|b\|$.

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There is an interesting article showing the equivalence of them; see Equivalency of Cauchy-Schwarz and Bessel Inequalities, The Mathematical Intelligencer December 2012, Volume 34, Issue 4, pp 2-3

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