proof of the inequality?

Question

prove that $$\frac{|x+y|}{1+|x+y|}\le\frac{|x|}{1+|x|}+\frac{|y|}{1+|y|} $$
To solve this I assumed if $x,y$ are of opposite signs and $x\gt y$ then $|x+y|\le |x|$. we get $$\frac{|x|}{1+|x|}\ge\frac{|x+y|}{1+|x+y|}$$ how to proceed further.. as answered by hans to prove that the function $$f(x)=\frac{x}{1+x}$$ is monotonous ,$y\ge x$ then $$\frac{1}{1+y}\le\frac{1}{1+x}$$ $$1-\frac{1}{1+y}\ge 1-\frac{1}{1+x}$$ $$\frac{y}{1+y}\ge\frac{x}{1+x}$$ further $|x+y|\le|x|+|y|$ and using the above inequality One gets

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

See also: Prove $\frac{|a+b|}{1+|a+b|}<\frac{|a|}{1+|a|}+\frac{|b|}{1+|b|}$. and Is $\frac{\vert x+y\vert}{1+\vert x+y\vert}\leq\frac{\vert x\vert}{1+\vert x\vert}+\frac{\vert y\vert}{1+\vert y\vert}$ true?. (And several other posts.) — Martin Sleziak, Feb 20 '24 at 14:07

Hans Engler · Answer 1 · 2016-09-07T13:30:19.227

5

This can be done in three steps:

Prove or quote that $|x+y| \le |x| + |y|$.
Prove that $f(z) = \frac{z}{z+1}$ is increasing on $[0,\infty)$. This can be done with Calculus or with an algebra step.
Prove that $\frac{a+b}{1+a+b} \le \frac{a}{1+a} + \frac{b}{1+b}$ for all $a, \, b \ge 0$. That is also an algebra step.

Then put these three arguments together to make a proof.

edited Sep 07 '16 at 13:30

answered Sep 07 '16 at 13:22

Hans Engler

16,092

1

It doesn't require calculus to show that $z\gt w\gt0$ implies $z/(z+1)\gt w/(w+1)$. – Barry Cipra Sep 07 '16 at 13:28
Indeed, all this can be done with algebraic manipulations. I'll change the post. – Hans Engler Sep 07 '16 at 13:29
yes it can be proved easily through inequalities @BarryCipra – Bogorovich Sep 07 '16 at 13:30

Matematleta · Answer 2 · 2016-09-07T14:17:10.317

3

All you need is the triangle inequality: $\vert x+y\vert \leq \vert x\vert +\vert y\vert$.

Now, if LHS=$0$, there is nothing to prove. Otherwise, we have

$$\frac{1}{\vert x\vert +\vert y\vert}\le\frac{1}{\vert x+y\vert }\Rightarrow 1+\frac{1}{\vert x\vert +\vert y\vert}\le 1+\frac{1}{\vert x+y\vert }\Rightarrow \frac{1}{1+\frac{1}{\vert x+y\vert }}\le \frac{1}{1+\frac{1}{\vert x\vert +\vert y\vert}}\Rightarrow \frac{\vert x+y\vert }{1+\vert x+y\vert }\le \frac{\vert x\vert +\vert y\vert }{1+\vert x\vert +\vert y\vert }=\frac{\vert x\vert}{1+\vert x\vert +\vert y\vert }+\frac{\vert y\vert }{1+\vert x\vert +\vert y\vert }\le \frac{\vert x\vert}{1+\vert x\vert }+\frac{\vert y\vert }{1+\vert y\vert }$$.

edited Sep 07 '16 at 14:17

answered Sep 07 '16 at 14:00

Matematleta

30,081

$\frac{|x+y|}{1+|x+y|}\le\frac{|x|}{1+|x|}+\frac{|y|}{1+|y|}$ is to be proved – Bogorovich Sep 07 '16 at 14:11
I added the extra step, which I thought was obvious. Thanks. – Matematleta Sep 07 '16 at 14:17
but how this trivial step is done,can you explain @chilango – Bogorovich Sep 07 '16 at 15:00
you are removing a positive number from each denominator, so each term increases – Matematleta Sep 07 '16 at 15:03
can you explain in detail @chilango – Bogorovich Sep 07 '16 at 15:05
Yes of course they will but there should be a way to prove it – Bogorovich Sep 07 '16 at 15:08
Are you joking? If $x,y>0$ then $x+y>y\Leftrightarrow 1/(x+y)<1/y$ – Matematleta Sep 07 '16 at 17:30

proof of the inequality?

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