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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,822

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

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

Hi,

We have

Adding those two would yield

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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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**mathdad****Member**- Registered: 2013-04-13
- Posts: 21

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

thus xz <= yz

and xz <= yz + xy

thus xz <= xy + yz

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,822

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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**mathdad****Member**- Registered: 2013-04-13
- Posts: 21

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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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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