How many proofs of the Cauchy-Schwarz inequality are there? Is there some kind of reference that lists all of these proofs?

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Here"s a simple proof:

$|vecxcdotvecy| leq |vecx||vecy| $

Substitute $|vecxcdotvecy| = |vecx||vecy|cos heta$

$| |vecx||vecy|cos heta |leq |vecx||vecy| $

Divide both sides by $|vecx||vecy|$

$ | cos heta| leq 1$

-Hey, I was looking for a "more serious" proof!

Then here you are!

Here"s another simple proof:

This is projecting a vector to lớn another one (Click the gif if it doesn"t load):


You drag its end in a line that is perpendicular to the other vector. Then multiply the length of the new vector with the old vector.

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Do you know what the multiplication is equal to? The dot sản phẩm of the vectors


When you project that vector, its norm (length) becomes lower - or stays the same if one of them is a scalar multiple of the other one.

^^ That was the proof. Think about it.

Source: $3$Blue$1$Brown

Wait, I look for a "really serious" proof!

Here you are.

Another proof:

Let $p(t)=||tvecy-vecx||^2$

As there"s an absolute value, it must be equal khổng lồ or bigger than $0$.

$p(t)=||tvecy-vecx||^2geq 0$

$p(t)=(tvecy-vecx)(tvecy-vecx)geq 0$

$p(t)=t^2(vecycdot vecy)-2t(vecxcdotvecy)+vecxcdot vecxgeq0$

Let"s substitute some things.

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$p(t)=t^2underbrace(vecycdot vecy)_colorbluelarge a+tunderbrace(-2vecxcdotvecy)_colorredlarge b+underbrace(vecxcdot vecx)_colorgreenlarge cgeq0$


Its minimum value must be $large frac-colorredb2colorbluea$

Substituting $large t= frac-colorredb2colorbluea$




Forget the $large p(t)$ function side (LHS)


Multiply by $large 4colorbluea$



$4colorblueacolorgreencgeq colorredb^2$


$p(t)=t^2underbrace(vecycdot vecy)_colorbluelarge a+tunderbrace(-2vecxcdotvecy)_colorredlarge b+underbrace(vecxcdot vecx)_colorgreenlarge cgeq0$

$4colorblue(vecycdot vecy)colorgreen(vecxcdot vecx)geq colorred(-2vecxcdotvecy)^2$

Using the identity $large vecvcdotvecv=||vecv||^2$

$4colorbluecolorgreengeq colorred(-2vecxcdotvecy)^2$

Using the identity $(f(x))^2=(|f(x)|)^2$ (where $f(x)inkumeR$)

$4colorbluecolorgreengeq colorred-2vecxcdotvecy^2$

As the both sides are not negative, you can square root both sides.

$2colorbluecolorgreengeq colorred$

$2colorbluecolorgreengeq colorredvecxcdotvecy$

$largecolorbluecolorgreengeq colorred$