About question 26, chapter 9, GRE 5 lb. book

[franalej]

About the answer to Chapter 9, question 26 on pages 398-399 of the GRE 5 lb book, please could you elucidate the reasoning of the following:

Since, x and y are reciprocals and the minimum value of x+y=2 and when plugging into the inequality for z:
if
z < 1-(x+y)
x < 1 - (at least 2)
then z < at most -1.
and therefore the answer is "B is greater".

Could you please explicate the relationship between "at least" 2 and then "at most" -1? Is there a number principle behind going from "at least" to "at most"?

Thanks,

-F

[gmbd97]

The question is :

x and y are positive numbers such that x+y+z<1 and xy=1

Quantity A =Z
Quantity B =-1


Lets take, x=1
so y must be 1 too!
now lets put x=1, y=1 in x+y+z<1
1+1+z<1
2+z <1
z <1-2
z<-1
So B must be greater!!

Now your question:
now if x=2, then y=1/2 ,so x+y=2.5
if x=3, then y=1/3 ,so x+y=3.33
so we can see that minimum value is when x=1, y=1, that is
x+y=2
so now,
x+y+z<1
z<1-(x+y)
z<1-(2.5) (when x=2,y-1/2)
z<-1.5
again, when x=3,y=1/3
z<1-(x+y)
z<1-(3.33)
z<-2.33
so we can see that our range is maximum when x=1, y=1, that's why they mentioned:z<-1(at most)

[tommywallach]

Thanks for rocking my job, gmb! That was perfect!

-t

[franalej]

I wonder if one could derive a definition that says that if we add a number with this characteristic 0<x<1, and its reciprocal, the result will be above 2.

Is this axiom true?

Thanks for your help!

-f

Thanks for rocking my job, gmb! That was perfect!

-t

[franalej]

Thanks for your explanation. It was helpful!.

-f

The question is :

x and y are positive numbers such that x+y+z<1 and xy=1

Quantity A =Z
Quantity B =-1


Lets take, x=1
so y must be 1 too!
now lets put x=1, y=1 in x+y+z<1
1+1+z<1
2+z <1
z <1-2
z<-1
So B must be greater!!

Now your question:
now if x=2, then y=1/2 ,so x+y=2.5
if x=3, then y=1/3 ,so x+y=3.33
so we can see that minimum value is when x=1, y=1, that is
x+y=2
so now,
x+y+z<1
z<1-(x+y)
z<1-(2.5) (when x=2,y-1/2)
z<-1.5
again, when x=3,y=1/3
z<1-(x+y)
z<1-(3.33)
z<-2.33
so we can see that our range is maximum when x=1, y=1, that's why they mentioned:z<-1(at most)

[tommywallach]

Hey Franalej,

I suppose you could, but it wouldn't be helpful. There's no need for extra axioms than the classics. This one will never come up again on a test, I promise you. : )

-t