# Absolute value...

## Homework Statement

X and Y 2 real numbers / |x| <1 and |y|<1
Prove that |x+y|<|xy+1|

## 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

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phinds
Gold Member
2019 Award
I couldn't prove that |xy+1| >2
That's good because
(1) It's not true
(2) It's not at all what you have been asked to prove.

PeroK
Homework Helper
Gold Member

## Homework Statement

X and Y 2 real numbers / |x| <1 and |y|<1
Prove that |x+y|<|xy+1|

## 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
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.

Filip Larsen
Gold Member
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 Mark44
Mentor
Note that you have, essentially, three cases:

1) ##x, y \ge 0##
2) ##x, y \le 0##
3) ##x \le 0, \ y \ge 0##
.
And this one?
4) ##x \ge 0, \ y \le 0##?

PeroK
Homework Helper
Gold Member
And this one?
4) ##x \ge 0, \ y \le 0##?
It's symmetric in ##x## and ##y##. That's just the same as case 3).

phinds
Gold Member
2019 Award
It's symmetric in ##x## and ##y##. That's just the same as case 3).
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.

PeroK
Homework Helper
Gold Member
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" in and then it does become just one condition.
That would be fair enough if I hadn't said "essentially". Formally, I would have said wlog (without loss of generality).

phinds
Gold Member
2019 Award
That would be fair enough if I hadn't said "essentially". Formally, I would have said wlog (without loss of generality).
A fair point. I hadn't notice that.

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

haruspex
Homework Helper
Gold Member
I think I got it?
Probably - I haven't checked the details - but there is a much easier way.
Note that |a|>|b| if and only if a2>b2.

• Filip Larsen
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

haruspex
Homework Helper
Gold Member
Yes I know i tried it out first
Would you mind posting that attempt?

haruspex
Homework Helper
Gold Member
Would you mind posting that attempt?
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.