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# proof of Cauchy-Schwarz inequality

If $a$ and $b$ are linearly dependent, we write $\boldsymbol{b}=\lambda\boldsymbol{a}$. So we get:

$\langle\boldsymbol{a},\lambda\boldsymbol{a}\rangle^{2}=\lambda^{2}\langle% \boldsymbol{a},\boldsymbol{a}\rangle^{2}=\lambda^{2}||\boldsymbol{a}||^{4}=||% \boldsymbol{a}||^{2}||\boldsymbol{b}||^{2}.$ |

So we have equality if $\boldsymbol{a}$ and $\boldsymbol{b}$ are linearly dependent. In the other case we look at the quadratic function

$||x\cdot\boldsymbol{a}+\boldsymbol{b}||^{2}=x^{2}||\boldsymbol{a}||^{2}+2x% \langle\boldsymbol{a},\boldsymbol{b}\rangle+||\boldsymbol{b}||^{2}.$ |

This function is positive for every real $x$, if $\boldsymbol{a}$ and $\boldsymbol{b}$ are linearly independent. Thus it has no real zeroes, which means that

$\langle\boldsymbol{a},\boldsymbol{b}\rangle^{2}-||\boldsymbol{a}||^{2}||% \boldsymbol{b}||^{2}$ |

is always negative. So we have:

$\langle\boldsymbol{a},\boldsymbol{b}\rangle^{2}<||\boldsymbol{a}||^{2}||% \boldsymbol{b}||^{2},$ |

which is the Cauchy-Schwarz inequality if $\boldsymbol{a}$ and $\boldsymbol{b}$ are linearly independent.

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## Comments

## a inequality question

i use cauchy-schwarz inequality to prove the follow inequality

Let (summation from i=1 to n) = S and X = multiply

S[a(suffix_i)]^(1/2) <or= {nXS[a(suffix_i)]}^(1/2)

it has no problem

the problem is "hence"

hence, prove that

1 + 1/2 + ... + 1/n <or= (2n-1)^(1/2)

i cannot do it

## Re: a inequality question

Are you sure that you quoted the original statement correctly? To me this does not look like something one would need C-S-inequality for as it is trivial.

--

"Do not meddle in the affairs of wizards for they are subtle and quick to anger."

## Re: a inequality question

Cases n=1 and n=2 are trivial to check. Now, sum 1/k <= sqrt( n sum 1/k^2) < sqrt(n pi^2/6) < sqrt (2n-1) as n>2. Summation index is k and it goes from 1 to n.