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    Show that:

    a^2+b^2+c^2 {\ge} ab+bc+ca
    for all positive integers a,b,c

    My proof:

    Following the conditions we have that:

    (a+b+c)^2{\ge}0

    a^2+b^2+c^2+2ab+2ac+2bc{\ge}0

    a^2+b^2+c^2{\ge}-2ab-2ac-2bc

    a^2+b^2+c^2{\ge}-2(ab+ac+bc)

    \frac{a^2+b^2+c^2}{2}{\ge}-(ab+ac+bc)

    Since we have that:

    2>-1

    We can use that in our original inequality:

    \frac{(a^2+b^2+c^2)*2}{2}{\ge}-(ab+ac+bc)*(-1)

    Finally:

    a^2+b^2+c^2 {\ge} ab+bc+ca

    Q.E.D.

    My questions is if I can use 2>-1 to cancel out the 2 on the LHS and make the RHS positive.

    Thanks.
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    (Original post by chrypton)
    Show that:

    a^2+b^2+c^2 {\ge} ab+bc+ca
    for all positive integers a,b,c

    My proof:

    Following the conditions we have that:

    (a+b+c)^2{\ge}0

    a^2+b^2+c^2+2ab+2ac+2bc{\ge}0

    a^2+b^2+c^2{\ge}-2ab-2ac-2bc

    a^2+b^2+c^2{\ge}-2(ab+ac+bc)

    \frac{a^2+b^2+c^2}{2}{\ge}-(ab+ac+bc)

    Since we have that:

    2>-1

    We can use that in our original inequality:

    \frac{(a^2+b^2+c^2)*2}{2}{\ge}-(ab+ac+bc)*(-1)

    Finally:

    a^2+b^2+c^2 {\ge} ab+bc+ca

    Q.E.D.

    My questions is if I can use 2>-1 to cancel out the 2 on the LHS and make the RHS positive.

    Thanks.
    No you can't because when you multiply an inequality by a negative number, the inequality sign changes.
    I would go about this inequality very differently by noting that:
    a^2+b^2+c^2\geq ab+ac+cb

    \Leftrightarrow a^2+b^2+c^2\geq \frac{1}{2}((a+b+c)^2-2(a^2+b^2+c^2))

    \Leftrightarrow \frac{a^2+b^2+c^2}{3}\geq (\frac{a+b+c}{3})^2.

    At this point, note that Root-Mean Square \geq Arithmetic mean.
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    If you start with:

    2 x 10 > -1 x 30 (which it does)

    and cancel using 2 > -1

    you get 10 > 30
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    (Original post by Farhan.Hanif93)
    No you can't because when you multiply an inequality by a negative number, the inequality sign changes.
    Ah, yes of course. Didn't think of that.
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    (Original post by chrypton)
    Ah, yes of course. Didn't think of that.
    Try something like (a-b)^2 + (b-c)^2 + (a-c)^2 \geq 0
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    (Original post by chrypton)
    Ah, yes of course. Didn't think of that.
    (Original post by Serano)
    Try something like (a-b)^2 + (b-c)^2 + (a-c)^2 \geq 0
    Read my edited first post, I'm fairly sure that's the easiest/quickest way (i.e. AM-QM inequality)
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    (Original post by Serano)
    Try something like (a-b)^2 + (b-c)^2 + (a-c)^2 \geq 0
    That's very elegant! Thanks
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    (Original post by Farhan.Hanif93)
    No you can't because when you multiply an inequality by a negative number, the inequality sign changes.
    I would go about this inequality very differently by noting that:
    a^2+b^2+c^2\geq ab+ac+cb

    \Leftrightarrow a^2+b^2+c^2\geq \frac{1}{2}((a+b+c)^2-2(a^2+b^2+c^2))

    \Leftrightarrow \frac{a^2+b^2+c^2}{3}\geq (\frac{a+b+c}{3})^2.

    At this point, note that Root-Mean Square \geq Arithmetic mean.
    Ok, but you can't start out with what you are going to show?
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    (Original post by chrypton)
    Ok, but you can't start out with what you are going to show?
    Of course you can. All I did was rearrange it to be in a form that definitely is true, and since that is true (the AM-QM bit), what you started with must be true. The implication goes both ways in this case.
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    (Original post by chrypton)
    Ok, but you can't start out with what you are going to show?
    In general, you are right. You should work from 'given' towards 'to prove'.

    Because Farhan's method uses double-implication signs <=> at all stages his proof's OK.
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    Thank you!
 
 
 
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