Inequality question.

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Dec 6th 2009, 04:53 PM
scorpion007

Inequality question.

I know this isn't strictly a Pre-Uni question, but it only involves basic algebra as far as I can tell.

Given the following picture:
http://img192.imageshack.us/img192/6616/math15.png
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I don't understand why $(x_1-x_2)^2 \ge 0 \implies P(n)~ \text{true for} ~n=2$ .

I can see how they derived the simpler inequality through simple algebraic manipulation, but don't understand why it means the statement holds for n=2.
Dec 6th 2009, 05:37 PM
TheEmptySet

Quote:

Originally Posted by scorpion007

I know this isn't strictly a Pre-Uni question, but it only involves basic algebra as far as I can tell.

Given the following picture:
http://img192.imageshack.us/img192/6616/math15.png
---------
I don't understand why $(x_1-x_2)^2 \ge 0 \implies P(n)~ \text{true for} ~n=2$ .

I can see how they derived the simpler inequality through simple algebraic manipulation, but don't understand why it means the statement holds for n=2.

Expand it out to get

$(x_1-x_2)^2=x_1^2+x_2^2-2x_1x_2 \ge 0 \iff x_1^2+x_2^2\ge 2x_1x_2 \ge x_1x_2$

Which is the statement of $P(2)$
Dec 6th 2009, 06:10 PM
scorpion007

Quote:

Which is the statement of P(2)

Isn't the statement of P(2):

$x_1 x_2 \le \frac{(x_1+x_2)^2}{4} $ ?

If I understand correctly, you've said that

$x_1 x_2 \le x_1^2 + x_2^2$

is the statement of P(2). Are they really the same?
Dec 6th 2009, 06:35 PM
TheEmptySet

Quote:

Originally Posted by scorpion007

Isn't the statement of P(2):

$x_1 x_2 \le \frac{(x_1+x_2)^2}{4} $ ?

If I understand correctly, you've said that

$x_1 x_2 \le x_1^2 + x_2^2$

is the statement of P(2). Are they really the same?

You are correct so let me do this again.

$(x_1+x_2)^2-4x_1x_2=(x_1^2-x_2^2) \ge 0 \implies$

$(x_1+x_2)^2-4x_1x_2 \ge 0 \iff (x_1+x_2)^2 \ge 4x_1x_2$

$\frac{(x_1+x_2)^2}{4} \ge x_1x_2$
Dec 6th 2009, 07:12 PM
scorpion007

Ah, perfect! thanks!
Dec 6th 2009, 07:33 PM
scorpion007

Actually, that raises a question:

To get to

$ (x_1+x_2)^2-4x_1x_2 \ge 0 $

We presumably would have started from

$ x_1 x_2 \le \frac{(x_1+x_2)^2}{4} $

and multiplied both sides by 4, then subtracted $4x_1x_2$ from both sides, right?

So haven't we just proven that $P(2) \iff P(2)$ . I.e. circular reasoning?
Dec 6th 2009, 08:21 PM
TheEmptySet

Quote:

Originally Posted by scorpion007

Actually, that raises a question:

To get to

$ (x_1+x_2)^2-4x_1x_2 \ge 0 $

We presumably would have started from

$ x_1 x_2 \le \frac{(x_1+x_2)^2}{4} $

and multiplied both sides by 4, then subtracted $4x_1x_2$ from both sides, right?

So haven't we just proven that $P(2) \iff P(2)$ . I.e. circular reasoning?

No just start with this

$(x_1-x_2)^2=x_1^2-2x_1x_2+x_2^2$

Now add and subtract $2x_1x_2$ to get

$x_1^2-2x_1x_2+x_2^2+2x_1x_2-2x_1x_2=(x_1^2+2x_1x_2+x_2^2)-4x_1x_2=(x_1+x_2)^2-4x_1x_2$
Dec 6th 2009, 08:28 PM
scorpion007

Right, but in the original image I posted, they don't start with $(x_1-x_2)^2$ , but rather from $(x_1+x_2)^2-4x_1x_2$ and then get to $(x_1-x_2)^2$ .

And it seems like to get to $(x_1+x_2)^2-4x_1x_2$ , they would have started with the definition of P(2).
Dec 7th 2009, 05:36 AM
TheEmptySet

Quote:

Originally Posted by scorpion007

Right, but in the original image I posted, they don't start with $(x_1-x_2)^2$ , but rather from $(x_1+x_2)^2-4x_1x_2$ and then get to $(x_1-x_2)^2$ .

And it seems like to get to $(x_1+x_2)^2-4x_1x_2$ , they would have started with the definition of P(2).

The equality is and if and only if statement. I.e all of the steps are reversible. So it you start with

$(x_1+x_2)^2-4x_1x_2$ and expand it out, collect like terms and factor you will end up with

$(x_1+x_2)^2-4x_1x_2=(x_1-x_2)^2$