Proof reviewer needed

Question

Originally I wanted to elaborate one idea of mine in my answer to this one question: How to prove that $\frac{|x+y+z|}{1+|x+y+z|} \le \frac{|x|}{1+|y|+|z|}+\frac{|y|}{|1+|x|+|z|}+\frac{|z|}{1+|x|+|y|}$

I'm trying to 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!

Answer 1

Let $|x_i|=a_i$ and $\sum\limits_{i=1}^{n}a_i=S.$

Thus, since by triangle inequality $$\frac{|\sum\limits_{i=1}^nx_i|}{1+|\sum\limits_{i=1}^nx_i|}=1+\frac{|\sum\limits_{i=1}^nx_i|}{1+|\sum\limits_{i=1}^nx_i|}-1=1-\frac{1}{1+|\sum\limits_{i=1}^nx_i|}\leq$$ $$\leq1-\frac{1}{1+\sum\limits_{i=1}^n|x_i|}=1-\frac{1}{1+S}=\frac{S}{1+S},$$ it's enough to prove that: $$\sum_{i=1}^n\frac{a_i}{1+S-a_i}\geq\frac{S}{1+S}.$$ Indeed, by C_S $$\sum_{i=1}^n\frac{a_i}{1+S-a_i}=\sum_{i=1}^n\frac{a_i^2}{a_i+Sa_i-a_i^2}\geq\frac{S^2}{S+S^2-\sum\limits_{i=1}^na_i^2}\geq\frac{S^2}{S+S^2}=\frac{S}{1+S}.$$

Answer 2

You can show that

$$ \frac { | \sum_i x_i | } { 1 + |\sum_i x_i| } \leq \sum_i \frac{|x_i| } { 1 + \sum_j |x_j| } < \sum_i \frac{|x_i| } { 1 + \sum_{j\neq i} |x_j| }.$$

See my solution for the full details.