Triangle Inequality question with fractions [duplicate]
Solution 1:
Because the function $f(x)=\frac{1}{1+\frac{1}{x}}=\frac{x}{1+x}$ is increasing, we have $$\frac{|a+b|}{1+|a+b|} = \frac{1}{1+\frac{1}{|a+b|}} \le \frac{1}{1+\frac{1}{|a|+|b|}}=\frac{|a|}{1+|a|+|b|}+\frac{|b|}{1+|a|+|b|}\le\frac{|a|}{1+|a|}+\frac{|b|}{1+|b|}$$ In the first inequality we have used the triangle inequality $|a+b|\le |a|+|b|$ and in the second inequality, the positivity of $|a|$ and $|b|$.
That is, equality holds if and only if $a=0$ or $b=0$ because otherwise the last inequality is strict.