Absolute value proof 5

Question

I've been asked to prove the following: $$\frac{|a+b|}{1+|a+b|} \le \frac{|a|}{1+|a|} + \frac{|b|}{1+|b|}$$

Does anyone got a tip for me?

Thank you in advance

Answer 1

$$\frac{|a|}{1+|a|}+\frac{|b|}{1+|b|}\geq\frac{|a|}{1+|a|+|b|}+\frac{|b|}{1+|a|+|b|}=$$ $$=\frac{|a|+|b|}{1+|a|+|b|}=1-\frac{1}{1+|a|+|b|}\geq1-\frac{1}{1+|a+b|}=\frac{|a+b|}{1+|a+b|}$$

Answer 2

Hint:

The function

$$f(x)=\frac{x}{1+x}\, \qquad (x>0)$$

is monotonically increasing because

$$f'(x)=\frac{1}{(1+x)^2}>0.$$

Hence by virtue of the following

$$|a+b|<|a|+|b|$$

One can say...what?