How to prove that $\frac{|x+y+z|}{1+|x+y+z|} \le \frac{|x|}{1+|y|+|z|}+\frac{|y|}{|1+|x|+|z|}+\frac{|z|}{1+|x|+|y|}$

Question

I need to prove $\frac{|x+y+z|}{1+|x+y+z|} \le \frac{|x|}{1+|y|+|z|}+\frac{|y|}{1+|x|+|z|}+\frac{|z|}{1+|x|+|y|}$. I've tried to use triangle inequality or to explore the form of $(a+b+c)^2$ but it won't get me anywhere. I would be grateful for some suggestions.

Answer 1

as $|x+y+z|\le |x|+|y|+|z|$

let $|x|=a,|y|=b,|z|=c$ it suffices to prove $$\sum \frac{a}{1+b+c}\ge \sum \frac{a+b+c}{1+a+b+c}$$
indeed by C-S/titu's lemma; $$\sum \frac{a}{1+b+c}=\sum \frac{a^2}{a+ba+ca}\ge \frac{{(a+b+c)}^2}{a+b+c+2(ab+bc+ca)}\ge \frac{a+b+c}{1+a+b+c}$$

Here we used $$\frac{2ab+2bc+2ca}{a+b+c}\le a+b+c$$ which is just $a^2+b^2+c^2\ge 0$

Answer 2

We will prove a stronger statement.

Lemma: $f(x) = \frac{x}{1+x}$ is an increasing function on $ x \geq 0$.
This is obvious by setting $ f(x) = 1 - \frac{1}{1+x}$, which is increasing on $ x \geq -1$.

Lemma: $ |x+y+z| \leq |x| + |y| + |z| $.
This is obvious by the basic properties of absolute values.

Corollary:

$$ \frac{ |x+y+z| } { 1 + |x+y+z|} \leq \frac{|x|+|y|+|z| } { 1 + |x| + |y| + |z| } \leq \sum \frac{ |x| } { 1 + |y| + |z| }$$

Answer 3

Let's check whether: $$S(N) = \frac{|\sum_{i=0}^{N} x_i|}{1+|\sum_{i=0}^{N} x_i|}\le \sum_{i=0}^{N} \frac{|x_i|}{1+ |\sum_{j=0;j \ne i}^{N} x_j|}$$ for all natural numbers $N$ and all natural numbers $i<N+1$ is true.

The inequation $\frac{|x+y+z|}{1+|x+y+z|} \le \frac{|x|}{1+|y|+|z|} + \frac{|y|}{1+|x|+|z|}+\frac{|z|}{1+|x|+|y|}$ is the special case for $i=2$ with $x=x_0, y=x_1, z=x_2$

For $N=0$ we have $\frac{|x_0|}{1+|x_0|} \le \frac{|x_0|}{1+ 0}$ for all $x_0$

Let's suppose for a certain $N$ that we have $$S(N) = \frac{|\sum_{i=0}^{N} x_i|}{1+|\sum_{i=0}^{N} x_i|} \le \sum_{i=0}^{N} \frac{|x_i|}{1+ |\sum_{j=0;j \ne i}^{N} x_j|}$$ for all natural numbers $N$ and all natural numbers $i<N+1$.

and prove this inequality for $S(N+1)$ which means let's prove the following:

$$S(N+1) \le \sum_{i=0}^{N+1} \frac{|x_i|}{1+ |\sum_{j=0;j \ne i}^{N+1} x_j|}$$ all natural numbers $i<N+2$.

We first will use $\frac{|a+b|}{1+|a+b|} \le \frac{|a|}{1+|a|} + \frac{|b|}{1+|b|}$ like proved in this link:Prove $\frac{|a+b|}{1+|a+b|}<\frac{|a|}{1+|a|}+\frac{|b|}{1+|b|}$.

for $a = x_{N+1}$ and $ b= \sum_{i=0}^{N} x_i$

$$\frac{|x_{N+1}+\sum_{i=0}^{N} x_i|}{1+|x_{N+1}+\sum_{i=0}^{N}x_i|} \le \frac{|x_{N+1}|}{1+|x_{N+1}|} + \frac{|\sum_{i=0}^{N} x_i|}{1+|\sum_{i=0}^{N} x_i|}$$

means that

$$\frac{|x_{N+1}+\sum_{i=0}^{N} x_i|}{1+|x_{N+1}+\sum_{i=0}^{N}x_i|} \le \frac{|x_{N+1}|}{1+|x_{N+1}|} + S(N)$$

means that $$ \frac{|x_{N+1}+\sum_{i=0}^{N} x_i|}{1+|x_{N+1}+\sum_{i=0}^{N}x_i|} \le \frac{|x_{N+1}|}{1+|x_{N+1}|} + \sum_{i=0}^{N} \frac{|x_i|}{1+ |\sum_{j=0;j \ne i}^{N} x_j|}$$

means that

$$S(N+1) \le \sum_{i=0}^{N+1} \frac{|x_i|}{1+ |\sum_{j=0;j \ne i}^{N+1} x_j|}$$

Proved!