Prove: |x+y|<|xy+1| for |x|,|y|<1

[liluiass]

Homework Statement

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

Homework Equations

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
Please help

 

[phinds]

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 Statement

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

Homework Equations

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
Please help

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]

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]

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]

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]

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]

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]

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.

 

[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

 

[haruspex]

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.

 

[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

 

[haruspex]

Yes I know i tried it out first

Would you mind posting that attempt?

 

[haruspex]

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.