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## #1 2013-04-09 18:30:43

Nehushtan
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## #2 2013-04-09 20:15:51

anonimnystefy
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From: Harlan's World
Registered: 2011-05-23
Posts: 16,037

### Re: Nehushtans challenge problems

#1 is not true. Try x=y=z=-2.

Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.

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## #3 2013-04-10 01:21:32

Nehushtan
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Why is it wrong?

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## #4 2013-04-15 02:06:41

gAr
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Registered: 2011-01-09
Posts: 3,479

### Re: Nehushtans challenge problems

Hi,

We have

"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense"  - Buddha?

"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."

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## #5 2013-04-15 02:22:03

Nehushtan
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### Re: Nehushtans challenge problems

Hi gAr.

You apparently assumed
,
,
to be positive. The result has to work for all real numbers, not just positive ones.

Last edited by Nehushtan (2013-04-15 02:25:42)

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## #6 2013-04-15 08:28:33

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Registered: 2013-04-13
Posts: 21

### Re: Nehushtans challenge problems

x <= y <= z implies Y >= x

thus xz <= yz

and  xz <= yz + xy

thus xz <= xy + yz

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## #7 2013-04-15 10:00:36

anonimnystefy
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From: Harlan's World
Registered: 2011-05-23
Posts: 16,037

### Re: Nehushtans challenge problems

That only works when all three numbers are positive.

Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.

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## #8 2013-04-15 11:54:55

Member
Registered: 2013-04-13
Posts: 21

### Re: Nehushtans challenge problems

Case 1: (same proof as my original answer)
assume x > 0
then, since x<=y<=z, and x > 0, then y > 0 and z > 0
Y >= x, and xz <= yz
and, xz <= yz + xy
thus xz <= xy + yz

Case 2:
assume x = 0,
then, since x<=y<=z, and x = 0, then y >= 0 and z >= 0
xz = 0, xy = 0, and yz >= 0
thus xz <= xy + yz

Case 3:
Assume y = 0,
then, since x<=y<=z, and y = 0, then x <= 0 and z >= 0
thus xz <= 0, xy = 0, and yz = 0
thus, xz <= xy + yz

Case 4:
Assume z = 0,
then, since x<=y<=z, and z = 0, then x <= 0 and y <= 0
xz = 0, xy >= 0, and yz = 0
thus, xz <= xy + yz

Case 5:
Assume z < 0,
then, since x<=y<=z, and z < 0, then x < 0 and y < 0
xz > 0, xy > 0, and yz > 0, (all are positive numbers)
since |x| >= |y| >= |z|
xy >= xz >= yz

Thus, xz <= xy
and, xz <= xy + yz

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## #9 2013-04-16 00:06:50

Nehushtan
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### Re: Nehushtans challenge problems

Splitting into cases is tedious. The short and sweet solution for #1 is

It looks like no-one is going to get #2 so I might as well post its solution as well.

New challenge problem:

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## #10 2013-04-24 10:36:57

Nehushtan
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### Re: Nehushtans challenge problems

Right, Ill set a simpler question this time.

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## #11 2013-05-03 11:49:24

Nehushtan
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