Proving the triangle inequality using the Cauchy-Schwartzs inequality

**s3a** · December 12th 2010, 11:56 AM

I googled and found something that seemed more complex than my Beginner's Linear Algebra course so can someone list me the steps in simplified form in order to prove this so that I can learn it?

Any help would be greatly appreciated!
Thanks in advance!

**skyking** · December 12th 2010, 12:20 PM

$||u+v||^2=||u||^2 +2Re(u,v)+||v||^2$ now you can use the inequality $Rez\leq|z|$ and the cauchy-schwarz inequality and from there it's trivial.

SK

**emakarov** · December 12th 2010, 12:31 PM

A general way to derive the triangle from the Cauchy–Schwarz inequality is shown in Wikipedia. To apply this to simple linear algebra, define the inner product of two vectors $\vec{x}=(x_1,\dots,x_n)$ and $\vec{y}=(y_1,\dots,y_n)$ , denoted by $\langle\vec{x},\vec{y}\rangle$ , to be $x_1y_1+\dots+x_ny_n$ . Also, let $||\vec{x}||=\sqrt{x_1^2+\dots+x_n^2}$ . Then $||\vec{x}||^2=\langle\vec{x},\vec{x}\rangle$ . It is easy to check that $\langle\vec{x}+\vec{y},\vec{z}\rangle=\langle\vec{ x},\vec{z}\rangle+\langle\vec{y},\vec{z}\rangle$ and $\langle\vec{x},\vec{y}\rangle=\langle\vec{y},\vec{ x}\rangle$ .

The Cauchy–Schwarz inequality then says that $|\langle\vec{x},\vec{y}\rangle| \leq \|\vec{x}\| \cdot \|\vec{y}\|$ , while the triangle inequality says that $||\vec{x}+\vec{y}||\le ||\vec{x}||+||\vec{y}||$ .

**Plato** · December 12th 2010, 01:16 PM

Originally Posted by s3a

I googled and found something that seemed more complex than my Beginner's Linear Algebra course so can someone list me the steps in simplified form in order to prove this so that I can learn it?

There is no point in our just guessing.
Why not tell us what your text material has to say about this?

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