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 $\displaystyle (x_1-x_2)^2 \ge 0 \implies P(n)~ \text{true for} ~n=2$.

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.

Quote:

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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 $\displaystyle (x_1-x_2)^2 \ge 0 \implies P(n)~ \text{true for} ~n=2$.

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.

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Expand it out to get

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

Which is the statement of $\displaystyle P(2)$

Quote:

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Which is the statement of P(2)

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Isn't the statement of P(2):

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

If I understand correctly, you've said that

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

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

Quote:

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Originally Posted by scorpion007

Isn't the statement of P(2):

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

If I understand correctly, you've said that

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

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

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You are correct so let me do this again.

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

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

$\displaystyle \frac{(x_1+x_2)^2}{4} \ge x_1x_2$

Ah, perfect! thanks!

Actually, that raises a question:

To get to

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

We presumably would have started from

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

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

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

Quote:

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Originally Posted by scorpion007

Actually, that raises a question:

To get to

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

We presumably would have started from

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

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

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

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No just start with this

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

Now add and subtract $\displaystyle 2x_1x_2$ to get

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

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

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

Quote:

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Originally Posted by scorpion007

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

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

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

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

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