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Thread: Inequality question.

  1. #1
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    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:

    ---------
    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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  2. #2
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    Quote Originally Posted by scorpion007 View Post
    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:

    ---------
    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.
    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 $\displaystyle P(2)$
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  3. #3
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    Which is the statement of P(2)
    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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    Quote Originally Posted by scorpion007 View Post
    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?
    You are correct so let me do this again.

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

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

    $\displaystyle \frac{(x_1+x_2)^2}{4} \ge x_1x_2$
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    Ah, perfect! thanks!
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    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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    Quote Originally Posted by scorpion007 View Post
    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?
    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

    $\displaystyle x_1^2-2x_1x_2+x_2^2+2x_1x_2-2x_1x_2=(x_1^2+2x_1x_2+x_2^2)-4x_1x_2=(x_1+x_2)^2-4x_1x_2$
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  8. #8
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    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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  9. #9
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    Quote Originally Posted by scorpion007 View Post
    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).
    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

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