- #1

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## 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

- Thread starter liluiass
- Start date

- #1

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X and Y 2 real numbers / |x| <1 and |y|<1

Prove that |x+y|<|xy+1|

|x+y|<2

I couldn't prove that |xy+1| >2

And couldn't find a way to solve the problem

Please help

- #2

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That's good becauseI couldn't prove that |xy+1| >2

(1) It's not true

(2) It's not at all what you have been asked to prove.

- #3

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Note that you have, essentially, three cases:## 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

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

Filip Larsen

Gold Member

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Edit: I type way too slow

- #5

Mark44

Mentor

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And this one?Note that you have, essentially, three cases:

1) ##x, y \ge 0##

2) ##x, y \le 0##

3) ##x \le 0, \ y \ge 0##

.

4) ##x \ge 0, \ y \le 0##?

- #6

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It's symmetric in ##x## and ##y##. That's just the same as case 3).And this one?

4) ##x \ge 0, \ y \le 0##?

- #7

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

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

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

- #9

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

- #10

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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

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Probably - I haven't checked the details - but there is a much easier way.I think I got it?

Note that |a|>|b| if and only if a

- #12

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Thats said it is shorter probably if I didn't make any mistake...I though I absolutely did

- #13

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Would you mind posting that attempt?Yes I know i tried it out first

- #14

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There being no response, I'll post my full solution.Would you mind posting that attempt?

|x|<1, |y|<1

x

(1-x

1+x

1+2xy+x

(1+xy)

|1+xy|>|x+y|

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

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