Question

Please help!

Answer 1

How can this be solved by inequalities?

Answer 2

I hope this is correct.
\(\begin{array}{ccc}&x+y&>&0\\&(x+y)^2&>&0\\&x^2+y^2&>&-2xy\end{array}\)

Answer 3

that is correct

Answer 4

I am not sure whether it should be a \(\ge\) inequality or whether the sign changes when divided by \(xy\).

Answer 5

note that if y = x = 1 then the expression = 2

Answer 6

No

Answer 7

i'm pretty sure it can't but I'm not sure how to prove it

Answer 8

Is my approach wrong?

Answer 9

x and y are both positive integers so sign does not change when you divide by xy

Answer 10

It should be \(\Large\frac{x^2+y^2}{xy}\normalsize\ge2\) in your response. :3

Answer 11

yes you are right I made a mistake there.

Answer 12

So,
\(\Large\frac{x^2+y^2}{xy}\normalsize>-2\) ?

Answer 13

- i got really confused!!! lol

Answer 14

why -2?

Answer 15

it cannot equal -2 because right side must be positive

Answer 16

Me too. I know it's easy to get the answer by substituting some arbitrary positive values, but I am trying to solve it using inequalities.

Answer 17

Have I done a step wrong?

Answer 18

I got -2 by moving 2xy to RHS.

Answer 19

yes
thats not wrong but its doesn't help us answer the question - x^2 + y^2 cannot equal -2xy but it is greater than it.

Answer 20

x^2 + y^2
-------- >= 2 is correct
xy

so we can say that the original expression is greater than or equal to 2

Answer 21

It's easier than I thought!

Answer 22

Hmm, so it seems like it's not possible to solve it by taking \(x+y > 0\).

Answer 23

oh hold on that started as an assumption!!

Answer 24

we definitely know that it can equal 2
and if we put some values in like 2 and 3 than its > 2 but we haven't proved it yet.

Answer 25

can we show that x^2 + y^2
--------- cannot be less than 2?
xy

Answer 26

So you mean to show that \(x\) and \(y\) are positive integers by solving it the other way?

Answer 27

x and y must be positive integers yes
and I dont think we can fit positive integers to the above to fit that equation.

Answer 28

x^2 + y^2 < 2xy
(x + y)^ - 2xy < 2xy

(x + y)^2 < 4xy

Answer 29

if x = y = 1 we have value of 4 each side
and looks like no other positive integers would fit
but i guess thats not a proof.

Answer 30

Ohh! This question makes my mind ache. >_<

Answer 31

Yea!!

Answer 32

Rebecca, any idea? :3

Answer 33

i mean if you take values of y + x greater than 2 like 1 and 2 for instance then they will definitly not fit this inequality , The lowest possible values are x =1 and y =1 and that gives a value of 4 each side .-

Answer 34

i have to go now . But if I have a brainwave I'l get back to you

Answer 35

Hehe. Thanks for giving it a try. :)

Answer 36

@Mehek14 any idea?

Answer 37

My brain has not been able to learn this stuff yet x.x

Answer 38

Ahh it's ok. :D

Answer 39

you last line
x^2 + y^2 > -2xy is incorrect beacuse

x^2 + y^2 = (x + y)^2 - 2xy

so adding 2xy to both sides gives
x^2 + y^2 > 2xy

Answer 40

thats a bit odd though because x =1 and y = 1 are valid solutions so x^2 + y^2 = 2xy in that case.

Answer 41

The lowest possible value of 2xy is 2*1*1 = 2 since x and y are positive integers. So x^2 + y^2 cannot be less than 2.
We have seen that x^2 + y^2 can be greater than 2 - so that completes the proof.


So ii is the correct choice.

Answer 42

so it looks like you started correctly with x + y > 0.

- though i'm a bit mystified why x=1 and y=1 is a solution

Answer 43

If y and x have to be different then the expression will always be > 2.

Answer 44

I guess by saying x + y > 0 and squaring and expanding the binomial this implies that x and y must have different values in order to do this manipulation so X^2 + y^2 would be > 2xy.
But its obvious that x=1 and y = 1 is a solution so the correct answer must be (ii).

Answer 45

I'm happy with that.

Answer 46

Last line of my previous calculation is not wrong. Instead of taking \(x+y>0\), I should have taken \((x-y)^2\ge0\). This leads to the correct answer.

\(\begin{array}{ccc}&(x-y)^2&\ge&0\\&x^2-2xy+y^2&\ge&0\\&x^2+y^2&\ge&2xy\\&\Large\frac{x^2+y^2}{xy}&\normalsize\ge&2\end{array}\)

Thanks for your help @welshfella. :)

Answer 47

AH yes Good work

Answer 48

I had a bad day yesterday! I kept making silly mistakes!