Proving the triangle inequality using the Cauchy-Schwartzs inequality

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• Dec 12th 2010, 11:56 AM
s3a
Proving the triangle inequality using the Cauchy-Schwartzs inequality
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!
• Dec 12th 2010, 12:20 PM
skyking
$\displaystyle ||u+v||^2=||u||^2 +2Re(u,v)+||v||^2$ now you can use the inequality $\displaystyle Rez\leq|z|$ and the cauchy-schwarz inequality and from there it's trivial.

SK
• Dec 12th 2010, 12:31 PM
emakarov
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 $\displaystyle \vec{x}=(x_1,\dots,x_n)$ and $\displaystyle \vec{y}=(y_1,\dots,y_n)$, denoted by $\displaystyle \langle\vec{x},\vec{y}\rangle$, to be $\displaystyle x_1y_1+\dots+x_ny_n$. Also, let $\displaystyle ||\vec{x}||=\sqrt{x_1^2+\dots+x_n^2}$. Then $\displaystyle ||\vec{x}||^2=\langle\vec{x},\vec{x}\rangle$. It is easy to check that $\displaystyle \langle\vec{x}+\vec{y},\vec{z}\rangle=\langle\vec{ x},\vec{z}\rangle+\langle\vec{y},\vec{z}\rangle$ and $\displaystyle \langle\vec{x},\vec{y}\rangle=\langle\vec{y},\vec{ x}\rangle$.

The Cauchy–Schwarz inequality then says that $\displaystyle |\langle\vec{x},\vec{y}\rangle| \leq \|\vec{x}\| \cdot \|\vec{y}\|$, while the triangle inequality says that $\displaystyle ||\vec{x}+\vec{y}||\le ||\vec{x}||+||\vec{y}||$.
• Dec 12th 2010, 01:16 PM
Plato
Quote:

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?