Let a and b denote 2 2d vectors

[LCKurtz]

oh okay, so i get

2 <= (a₁²b₂² + a₂²b₁²) / a₁b₁a₂b₂

or if i move everything i get

0 <= a₁²b₂² + a₂²b₁² - 2a₁b₁a₂b₂

OK. Only the second expression has everything on the right side. So now you have to figure out why that is $\ge 0$.

 

[Mark44]

oh okay, so i get

2 <= (a₁²b₂² + a₂²b₁²) / a₁b₁a₂b₂

The above won't do you any good.

or if i move everything i get

0 <= a₁²b₂² + a₂²b₁² - 2a₁b₁a₂b₂

Or more suggestively, 0 <= a₁²b₂² - 2a₁b₁a₂b₂ + a₂²b₁²

There's something you need to do with this...

 

[ibaforsale]

OK. Only the second expression has everything on the right side. So now you have to figure out why that is $\ge 0$.

is it greater than 0 because the squared terms will always be greater?

 

[LCKurtz]

is it greater than 0 because the squared terms will always be greater?

How do you know that? Are you just guessing?

 

[ibaforsale]

my thinking is that squaring something will always be bigger than multiplying by 2

 

[Office_Shredder]

First, the things you are squaring aren't the same as the things you are multiplying by two, and secondly, (1/2)² < 2*(1/2) so the basic logic there doesn't work.

You need to do some algebra to re-write the right hand side.

 

[LCKurtz]

Squaring $1$ isn't bigger than multiplying it by $2$. Look, you need to do some algebraic manipulation with that expression to make it obvious that is is $\ge 0$. We have led you by the hand to this point and I don't see how we can give you that last step without having worked the whole problem for you.

As a matter of curiosity, what course are you in and what courses have you already had?

 

[ibaforsale]

i was thinking the algebra might include taking out common terms but there are none, can i get a hint as to what needs to happen

 

[Mark44]

We have led you by the hand to this point and I don't see how we can give you that last step without having worked the whole problem for you.

That's how I feel as well.