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Given $0\leq a\leq c+g$

and $0\leq b\leq d+f.$

I want to show that $\sqrt{( a^2+b^2)}\le\sqrt{(c^2+d^2)}+\sqrt{(f^2+g^2)}.$

I have tried working both ways, i.e. both building up from the given and assuming the result and working backwards, but can't figure out how to get rid of the extra terms obtained from squaring both sides.

Any tips? Perhaps I am missing some obvious property?

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Particular case in $\mathbb{R}^2$ of the discrete case of: – 1015 Jan 11 '13 at 2:14
up vote 6 down vote accepted

Triangle inequality on the points $(0, 0), (-c, -d), (g, f)$ shows that

$\sqrt{(c+g)^2 + (d+f)^2} \leq \sqrt{c^2 +d^2} + \sqrt{g^2+f^2}$

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I think you mean $\leq$ instead of $<$, but otherwise, good answer! +1 – Clayton Jan 11 '13 at 2:07
@Clayton Yes! Thanks. Not sure why I did that ... – Calvin Lin Jan 11 '13 at 2:10

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