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Use the triangle inequalities to prove that:

$$\frac{|a+b|}{1+|a+b|}\le\frac{|a|}{1+|a|} + \frac{|b|}{1+|b|}$$ for all $a,b\in\mathbb{R}$.

I've made it this far, but it doesn't seem to be helping much; what I end up with at the end isn't very useful:

enter image description here

Clearly I'm taking a wrong turn somewhere, but I'm not sure what else to try. Any advice?

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  • $\begingroup$ You aren't considering that $|a|+|b| \ge |a|$. So you are closer than you think. $\endgroup$
    – fleablood
    Feb 13, 2018 at 5:43

1 Answer 1

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$$\frac{|a+b|}{1+|a+b|}=1-\frac{1}{1+|a+b|}\leq1-\frac{1}{1+|a|+|b|}=$$ $$=\frac{|a|}{1+|a|+|b|}+\frac{|b|}{1+|a|+|b|}\leq\frac{|a|}{1+|a|}+\frac{|b|}{1+|b|}.$$

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  • $\begingroup$ Man, I should have known to try that considering the form it was written in. Thanks a bundle. $\endgroup$
    – EllipticalInitial
    Feb 13, 2018 at 5:38

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