Inequality with squares and square roots.

Question

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?

Answer 1

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

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