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Math Help - 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 (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 (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)
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  3. #3
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    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}<br />
?

    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?
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    Quote Originally Posted by scorpion007 View Post
    Isn't the statement of P(2):

    x_1 x_2 \le \frac{(x_1+x_2)^2}{4}<br />
?

    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
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  5. #5
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    Ah, perfect! thanks!
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  6. #6
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    Actually, that raises a question:

    To get to

    <br />
(x_1+x_2)^2-4x_1x_2 \ge 0<br />

    We presumably would have started from

    <br />
x_1 x_2 \le \frac{(x_1+x_2)^2}{4}<br />

    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?
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    Quote Originally Posted by scorpion007 View Post
    Actually, that raises a question:

    To get to

    <br />
(x_1+x_2)^2-4x_1x_2 \ge 0<br />

    We presumably would have started from

    <br />
x_1 x_2 \le \frac{(x_1+x_2)^2}{4}<br />

    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
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  8. #8
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    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).
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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 (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
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