how to prove the following inequality by using real analysis

Question

how to prove this inequality by real anlysis

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

I finally came with the following inequality. but I dont know what to do after that

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

Possible duplicate of Prove $\frac{|a+b|}{1+|a+b|}<\frac{|a|}{1+|a|}+\frac{|b|}{1+|b|}$. — Martin R, Feb 01 '19 at 10:00
Also: https://math.stackexchange.com/q/563491/42969, https://math.stackexchange.com/q/983043/42969 – all found with Approach0 — Martin R, Feb 01 '19 at 10:01

score 2 · Accepted Answer · answered Feb 01 '19 at 07:49

Note that the function $f(t) = \frac{t}{1+t}$ is increasing for $t\geq 0$.

So, using

$|x+y| \leq |x| + |y|$ and
$\frac{|x|}{1+|x|} \geq \frac{|x|}{1+|x|+|y|}$ and $\frac{|y|}{1+|y|} \geq \frac{|y|}{1+|x|+|y|}$

you get

$$\frac{|x|}{1+|x|} + \frac{|y|}{1+|y|} \geq \frac{|x| + |y|}{1+|x| +|y|} \stackrel{f\; increasing}{\geq}\frac{|x+y|}{1+|x+y|}$$

score 1 · Answer 2 · answered Feb 01 '19 at 07:45

1

Let $f(t):= \frac{t}{1+t}$ for $t \ge 0.$ Show that $f$ is increasing. Since $|x+y| \le |x|+|y|$, we get

$\frac{|x+y|}{1+|x+y|}=f(|x+y|) \le f(|x|+|y|)=\frac{|x|+|y|}{1+|x|+|y|}$.

Can you proceed ?

answered Feb 01 '19 at 07:45

Fred

score 0 · Answer 3 · answered Feb 01 '19 at 07:58

0

Just use the triangle inequality.

$$ \frac{|x+y|}{1+|x|+|y|} \leq \frac{|x|+|y|}{1+|x|+|y|}=\frac{|x|}{1+|x|+|y|}+\frac{|y|}{1+|x|+|y|}\leq \frac{|x|}{1+|x|}+\frac{|y|}{1+|y|} $$

answered Feb 01 '19 at 07:58

PierreCarre

Best answer +1. – Feb 01 '19 at 08:34
The task is to estimate $\frac{|x+y|}{1+|x+y|} $ and that is $\ge \frac{|x+y|}{1+|x|+|y|} $ ... – Martin R Feb 01 '19 at 10:04
You are right, I missread the initial post. – PierreCarre Feb 01 '19 at 13:14

3 Answers3