# Absolute value...

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1. Jul 17, 2016

### liluiass

1. The problem statement, all variables and given/known data
X and Y 2 real numbers / |x| <1 and |y|<1
Prove that |x+y|<|xy+1|

2. Relevant equations

3. The attempt at a solution
|x+y|<2
I couldn't prove that |xy+1| >2
And couldn't find a way to solve the problem

2. Jul 17, 2016

### phinds

That's good because
(1) It's not true
(2) It's not at all what you have been asked to prove.

3. Jul 17, 2016

### PeroK

Note that you have, essentially, three cases:

1) $x, y \ge 0$
2) $x, y \le 0$
3) $x \le 0, \ y \ge 0$

You could try those cases separately and see what you can do.

4. Jul 17, 2016

### Filip Larsen

Since, by definition, |x| = x when x ≥ 0 and |x| = -x when x < 0, a general method for "removing" absolute values is to divide each instance of an absolute value into two cases. Perhaps you can find a way to divide the relation you are trying to prove into such different cases? For instance, how must x and y relate to each other if you are to "remove" the absolute value on |x+y| ?

Edit: I type way too slow

5. Jul 17, 2016

### Staff: Mentor

And this one?
4) $x \ge 0, \ y \le 0$?

6. Jul 17, 2016

### PeroK

It's symmetric in $x$ and $y$. That's just the same as case 3).

7. Jul 17, 2016

### phinds

You're thinking one step ahead and realizing that it works out that way, but for slow people like me it's beneficial to state ALL the cases up front so I think Mark's statement is appropriate.

I MIGHT have thought to make one of the conditions just "x and y have different signs" and then it does become just one condition.

8. Jul 17, 2016

### PeroK

That would be fair enough if I hadn't said "essentially". Formally, I would have said wlog (without loss of generality).

9. Jul 17, 2016

### phinds

A fair point. I hadn't notice that.

10. Jul 19, 2016

### liluiass

I think I got it?
If x>0 and y>0
Then |x+y |= x+ y and | xy+1|=xy +1
We have to prove that x+y < xy +1
X-1 < xy -y
X-1 < y(x-1) / x-1<0
1>y which is true tgeb x+y <xy+1
--if x and y are negative
|x+y |= -x-y and xy +1>0 then |xy+1 |=xy +1
Let us prove again thet -x-y <xy +1
-x-xy<1+y
-x(1+y)< 1+y / -1<y<0 => y+1>0
-x<1< = > x>-1 which is true .
----case 3 if xis positif and y is negatif
0<x<1 , -1<y<0
-1<x+y<1 ,1+ xy<1
If x+y >0
We have to prove : x+y < xy +1
X-1< y(x-1) /x-1 <0
1> y true.
If x++y <0 ..we have to prove that -x-y <xy +1 ..
-x(1+y) <1+y ../ 1+y >0
So -x<1
X>-1 which is true

11. Jul 19, 2016

### haruspex

Probably - I haven't checked the details - but there is a much easier way.
Note that |a|>|b| if and only if a2>b2.

12. Jul 19, 2016

### liluiass

Yes I know i tried it out first and didn't knowif its alright to use the same method bec I couldn't remember how inequalities work at this point
Thats said it is shorter probably if I didn't make any mistake...I though I absolutely did

13. Jul 19, 2016

### haruspex

Would you mind posting that attempt?

14. Jul 21, 2016

### haruspex

There being no response, I'll post my full solution.
|x|<1, |y|<1
x2<1, y2<1
(1-x2)(1-y2)>0
1+x2y2>x2+y2
1+2xy+x2y2>x2+2xy+y2
(1+xy)2>(x+y)2
|1+xy|>|x+y|

Note that |x|>1, |y|>1 leads to the same result.