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\$

\$p(t)=colorblueat^2+colorredbt+colorgreencgeq0\$

Its minimum value must be \$large frac-colorredb2colorbluea\$

Substituting \$large t= frac-colorredb2colorbluea\$

\$p(frac-colorredb2colorbluea)=colorbluea(frac-colorredb2colorbluea)^2+colorredb(frac-colorredb2colorbluea)+colorgreencgeq0\$

\$p(frac-colorredb2colorbluea)=colorbluea(fraccolorredb^24colorbluea^2)+colorredb(frac-colorredb2colorbluea)+colorgreencgeq0\$

\$p(frac-colorredb2colorbluea)=fraccolorredb^24colorbluea+frac-colorredb^22colorbluea+colorgreencgeq0\$

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

\$fraccolorredb^24colorbluea+frac-colorredb^22colorbluea+colorgreencgeq0\$

Multiply by \$large 4colorbluea\$

\$colorredb^2-2colorredb^2+4colorblueacolorgreencgeq0\$

\$-colorredb^2+4colorblueacolorgreencgeq0\$

\$4colorblueacolorgreencgeq colorredb^2\$

De-substitute

\$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\$