# Let u and v be two vectors in RR^2 . The Cauchy-Schwarz inequality states that |u * v| <= |u||v| We are able to transform the above inequality so that it also shows us that |u+v| <= |u|+|v| But I cannot find a way to show that |u−v| <= |u|+|v| even though I now this has to be true. Any ideas?

Let u and v be two vectors in ${\mathbb{R}}^{2}$
The Cauchy-Schwarz inequality states that $|u·v|\le |u||v|$
We are able to transform the above inequality so that it also shows us that
$|u+v|\le |u|+|v|$
But I cannot find a way to show that $\mathbf{|}\mathbf{u}\mathbf{-}\mathbf{v}\mathbf{|}\mathbf{\le }\mathbf{|}\mathbf{u}\mathbf{|}\mathbf{+}\mathbf{|}\mathbf{v}\mathbf{|}$
even though I now this has to be true. Any ideas?
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ri1men4dp
$|u+v|\le |u|+|v|$
is shown already, so let's use this.
$|u-v|=|u+\left(-v\right)|\le |u|+|-v|=|u|+|v|$
and we're done.