Inequalities and absolute values.

Question

Prove that for every $x,y\in\mathbb{R},$ $$\dfrac{|x+y|}{1+|x+y|}\le\frac{|x|}{1+|x|}+ \frac{|y|}{1+|y|}.$$

I really don't know how to start. I thought to dividing it into 4 parts when $(x<0) \land (y>0)$, $(x>0)\land (y<0)$, $(x>0)\land (y>0)$, $(x<0)\land (y<0)$ but I didn't get much from this.

If you can do the $x>0$, $y>0$ case all the other cases will follow rapidly. — Angina Seng, Dec 01 '19 at 18:07
Possible duplicate of Prove $\frac{|a+b|}{1+|a+b|}<\frac{|a|}{1+|a|}+\frac{|b|}{1+|b|}$. — Martin R, Dec 01 '19 at 18:12
Also: https://math.stackexchange.com/q/983043/42969, https://math.stackexchange.com/q/2277665/42969, https://math.stackexchange.com/q/2386990/42969, https://math.stackexchange.com/q/698869/42969. — Martin R, Dec 01 '19 at 18:13

score 2 · Answer 1 · answered Dec 01 '19 at 18:15

The inequality also gives us

$$\dfrac{1}{\left| x + y \right|} \geq \dfrac{1}{\left| x \right| + \left| y \right|}.$$

Then,

$$1 + \dfrac{1}{\left| x + y \right|} \geq 1 + \dfrac{1}{\left| x \right| + \left| y \right|},$$

so that,

$$\dfrac{1}{1 + \frac{1}{\left| x + y \right|}} \leq \dfrac{1}{1 + \frac{1}{\left| x \right| + \left| y \right|}}.$$

Simple rearrangement of terms gives us

$$\dfrac{\left| x + y \right|}{1 + \left| x + y \right|} \leq \dfrac{\left| x \right|}{1 + \left| x \right| + \left| y \right|} + \dfrac{\left| y \right|}{1 + \left| x \right| + \left| y \right|}.$$

Since $\left| x \right| \geq 0$ and $\left| y \right| \geq 0$, your inequality now follows.

Inequalities and absolute values.

1 Answers1