Nehushtans challenge problems

Nehushtan · 2013-04-09 18:30:43

anonimnystefy · 2013-04-09 20:15:51

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

Nehushtan · 2013-04-10 01:21:32

Why is it wrong?

gAr · 2013-04-15 02:06:41

Hi,

We have

Adding those two would yield

Nehushtan · 2013-04-15 02:22:03

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)

mathdad · 2013-04-15 08:28:33

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

thus xz <= yz

and xz <= yz + xy

thus xz <= xy + yz

anonimnystefy · 2013-04-15 10:00:36

That only works when all three numbers are positive.

mathdad · 2013-04-15 11:54:55

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

Nehushtan · 2013-04-16 00:06:50

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:

Nehushtan · 2013-04-24 10:36:57

Right, Ill set a simpler question this time.

Nehushtan · 2013-05-03 11:49:24

Math Is Fun Forum

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