proof of Cauchy-Schwarz inequality for real numbers

The version of the Cauchy-Schwartz inequality we want to prove is

$\left(\sum_{k=1}^{n}a_{k}b_{k}\right)^{2}\leq\sum_{k=1}^{n}a_{k}^{2}\cdot\sum_% {k=1}^{n}b_{k}^{2},$

where the $a_{k}$ and $b_{k}$ are real numbers, with equality holding only in the case of proportionality, $a_{k}=\lambda b_{k}$ for some real $\lambda$ for all $k$ .

The proof is by direct calculation:

	$\displaystyle\sum_{k=1}^{n}a_{k}^{2}\cdot\sum_{k=1}^{n}b_{k}^{2}-\left(\sum_{k% =1}^{n}a_{k}b_{k}\right)^{2}$	$\displaystyle=\sum_{k,l=1}^{n}a_{k}^{2}b_{l}^{2}-a_{k}b_{k}a_{l}b_{l}$
		$\displaystyle=\sum_{k,l=1}^{n}\frac{1}{2}(a_{k}^{2}b_{l}^{2}+a_{l}^{2}b_{k}^{2% })-(a_{k}b_{l})(a_{l}b_{k})$
		$\displaystyle=\frac{1}{2}\sum_{k,l=1}^{n}(a_{k}b_{l})^{2}-2(a_{k}b_{l})(a_{l}b% _{k})+(a_{l}b_{k})^{2}$
		$\displaystyle=\frac{1}{2}\sum_{k,l=1}^{n}(a_{k}b_{l}-a_{l}b_{k})^{2}$
		$\displaystyle\geq 0.$

The above identity implies that the Cauchy-Schwarz inequality holds. Moreover, it is an equality only when

$a_{k}b_{l}-a_{l}b_{k}=0\quad\Longleftrightarrow\quad\frac{a_{k}}{b_{k}}=\frac{% a_{l}}{b_{l}}\text{ or }\frac{b_{k}}{a_{k}}=\frac{b_{l}}{a_{l}}\text{ or }a_{k% }=b_{k}=0,$

for all $k$ and $l$ . In other words, equality holds only when $a_{k}=\lambda b_{k}$ for all $k$ for some real number $\lambda$ .

Title	proof of Cauchy-Schwarz inequality for real numbers
Canonical name	ProofOfCauchySchwarzInequalityForRealNumbers
Date of creation	2013-03-22 14:56:38
Last modified on	2013-03-22 14:56:38
Owner	stitch (17269)
Last modified by	stitch (17269)
Numerical id	5
Author	stitch (17269)
Entry type	Proof
Classification	msc 15A63