# Prove that $\frac{|x+y|}{1+|x+y|}\leq\frac{|x|}{1+|x|}+\frac{|y|}{1+|y|}$ for any $x,y$ [duplicate]

Use mean value theorem to prove the inequality:

$$\frac{|x+y|}{1+|x+y|}\leq\frac{|x|}{1+|x|}+\frac{|y|}{1+|y|}\quad \forall\, x,y\in\mathbb{R}$$

I have no idea which function I should consider to apply the theorem. I tried $$\ln(|1+x|)$$, but this function is not defined at $$-1$$.

## marked as duplicate by Lord Shark the Unknown, mrtaurho, Parcly Taxel, Cesareo, YiFanMar 18 at 10:18

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• Maybe it would be better to start off with proving that $\frac{1}{1+|x+y|}\le\frac{1}{1+|x|}+\frac{1}{1+|y|}$ since you know the triangle inequality already holds. – Cameron Williams Nov 12 '13 at 5:51
• Try something along the lines of log (1 + |x|) $\pm$ log(1 + |kx|) where k is a constant that you are going to choose later, and will have something to do with what y you have. – Betty Mock Nov 12 '13 at 6:21

## 1 Answer

$\bf{Hint:}$ Suppose that $f:[0,\infty)\rightarrow[0,\infty)$ is a continuously differentiable function such that $f(0)=0$, $f'$ is non-negative and monotone decreasing. Then $s,t\geq 0$ $$0 \leq f(s+t) \leq f(s)+f(t)$$ and for $0\leq t_1\leq t$ $$0\leq f(t_1)\leq f(t_2).$$ Consider the function $$f(t) = \dfrac{t}{1+t}$$ and its derivatve $$f'(t) = \dfrac{1}{(1+t)^2}.$$

What you want to show is the sub-additive property: you can use this:

$$f(s+t) = \int_0^{s+t}f'(x) dx = \int_0^s f'(x) dx+\int_s^{s+t} f'(x)dx$$