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Math Help - Masked algebric inequality

  1. #1
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    Masked algebric inequality

    Prove that \sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2} >\sqrt{a^2+ac+c^2},if a,b,c>0.
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: Masked algebric inequality

    Quote Originally Posted by TheodorMunteanu View Post
    Prove that \sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2} >\sqrt{a^2+ac+c^2},if a,b,c>0.
    I'm not so sure, here it goes...

    I used Inequality of arithmetic and geometric means - Wikipedia, the free encyclopedia

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


    \sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2} \geq 2\sqrt{\sqrt{(a^2-ab+b^2)(b^2-bc+c^2)}}\geq 2\sqrt{\sqrt{(a^2-2ab+b^2)(b^2-2bc+c^2)}}=2\sqrt{\sqrt{(a-b)^2(b-c)^2}}=2\sqrt{(a-b)(b-c)}>2\sqrt{\frac{(a+c)^2}{4}}=\sqrt{(a+c)^2}>\sqrt  {a^2+ac+c^2}
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  3. #3
    Newbie Auri's Avatar
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    Re: Masked algebric inequality

    You could have also substituted a, b, and c with any number greater than 0, and see if the statement returned true.

    Something like.
    Whops sorry, remove this-

    What I meant though was substitute 1 for all three variables and see if the function is true.

    \sqrt{1^2-1*1+1^2}+\sqrt{1^2-1*1+1^2} >\sqrt{1^2+1*1+1^2}

    All though I could be completely wrong, so sorry if I am. Please correct.
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  4. #4
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    Re: Masked algebric inequality

    I was thinking that if we take OA=x,OB=y,OC=z with \angle AOB=60,\angle BOC=60\Righarrow AB=\sqrt{a^2-ab+b^2},BC=\sqrt{b^2-bc+c^2},AC=\sqrt{a^2+ac+c^2} and using the inequality of triangle in ABC we got the conclusion.
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  5. #5
    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: Masked algebric inequality

    Quote Originally Posted by Auri View Post
    You could have also substituted a, b, and c with any number greater than 0, and see if the statement returned true.

    Something like.
    Whops sorry, remove this-

    What I meant though was substitute 1 for all three variables and see if the function is true.

    \sqrt{1^2-1*1+1^2}+\sqrt{1^2-1*1+1^2} >\sqrt{1^2+1*1+1^2}

    All though I could be completely wrong, so sorry if I am. Please correct.

    You proved that is true for a=b=c=1, but you asked to show that the equality is true for all real positive numbers a,b,c.


    ------------------

    Now I see that I used the condition a>b>c>0 (I assuming that it is allowed... hmmm...)
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  6. #6
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    Re: Masked algebric inequality

    Also try to prove that \sqrt{x^2+y^2-\sqrt{3}xy}+\sqrt{y^2+z^2-yz} >\sqrt{x^2+z^2}
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