Inequality question (1 Viewer)

[Chris100]

Given that (x+y)^2>=4xy,
Prove that 1/x^2 +1/y^2>=4/(x^2+y^2)

 

[RealiseNothing]

Using the AM-GM inequality.

 

[QZP]

I don't understand your step from 2/xy to >= 4/(x^2 + y^2) :S

 

[RealiseNothing]

I don't understand your step from 2/xy to >= 4/(x^2 + y^2) :S

Flip the AM-GM.

 

[QZP]

Flip the AM-GM.

Ah right, thanks.

 

[Chris100]

What is AM and GM?

 

[QZP]

What is AM and GM?

Arithmetic mean of two numbers a,b: (a+b)/2
Geometric mean: root_ab

The AM-GM inequality says that (a+b)/2 >= root_ab

This is essentially what you are given

 

[Chris100]

Oh right
How did realise nothing apply that to the question?

 

[QZP]

Oh right
How did realise nothing apply that to the question?

It is quite easy to get to 2/xy with algebraic manipulation. Then he realised that:
(a+b)^2 /4 = ab
(a^2 + 2ab + b^2)/4 = ab
(a^2 + b^2)/4 + ab/2 = ab
(a^2 + b^2)/4 = ab/2
4/(a^2 + b^2) = 2/ab

Just apply that to the inequalities in the question (I'm lazy sorry)

 

[Chris100]

Oh wow that's clever!
What year do students usually learn the AM-GM inequality thereom? Not sure if I was just not listening in class or yr 12 theory

 

[asianese]

This is a 4U Question not a 3U one.

 

[panda15]

Oh wow that's clever!
What year do students usually learn the AM-GM inequality thereom? Not sure if I was just not listening in class or yr 12 theory

It comes up in 2U series and sequences, but I think it's pretty rare for the AM-GM inequality to appear outside of 4U.

 

[Trebla]

Using the AM-GM inequality.

There is a flaw here. Note that the AM-GM inequality



does NOT necessarily imply that



unless xy > 0. If xy < 0 then the argument breaks down.

 

[Trebla]

The approach I'm thinking of is basically using the knowledge that



in general. Simply due to the fact that

Since this is true for any real numbers x and y, then it must hold true for real numbers x² and y² which means that



Rearrange this inequality and the result follows without having to worry about negativity problems.

 

[anomalousdecay]

There is a flaw here. Note that the AM-GM inequality



does NOT necessarily imply that



unless xy > 0. If xy < 0 then the argument breaks down.

Well spotted trebla.

OP should check whether there are any conditions in the question.

 

[Chris100]

I didn't use AM-GM for x²+y²>=2xy,
I used the given (x+y)²>=4xy and if you expand the LHS and minus 2xy from both sides, you get x²+y²>=2xy

Therefore, (x²+y²)/x²y² >=2xy/x²y²