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.