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Derive the inequality

  1. Nov 23, 2006 #1
    For a triangle with sides a,b,c and its corresponding circle with radius R:



    [tex]\frac{a^2b^2}{c^2} +\frac{a^2c^2}{b^2}+\frac{b^2c^2}{a^2} \geq 9R^2 [/tex]
     
    Last edited: Nov 23, 2006
  2. jcsd
  3. Nov 23, 2006 #2

    Office_Shredder

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    What is the corresponding circle? Inscribed or circumscribed?
     
  4. Nov 23, 2006 #3
    Circumscribed of course (usually denoted by capital R).
     
  5. Nov 24, 2006 #4
    Anybody?
    I prooved it in a long and ugly way.
    I'd like to see ,if possible,an elegant proof of it.
     
  6. Nov 29, 2006 #5
    me too.
    and tehno, can you please post your proof or at least an outline of it?
     
  7. Dec 2, 2006 #6
    Nothing particularly smart.
    I used substitution:
    [tex]R=\frac{abc}{4P}[/tex]
    where P is area of triangle with sides a,b,c.
    I went to proove :
    [tex]\frac{ab}{c^2}+\frac{ac}{b^2}+\frac{bc}{a^2}>\frac{9}{4}[/tex]
    which I used along the way to proove the original inequality having on mind basic triangle inequality [tex]a+b>c[/tex].
    After lot of algebraic work I arrived at the original inequality.

    But can we somehow make a use something more elegant like a well known :
    [tex]\frac{1}{a^2+b^2+c^2}\geq \frac{1}{9R^2}[/tex]

    ?
     
  8. Dec 20, 2006 #7
    It can't be possible you have a proof of it becouse the inequality is invalid!
    Some things were odd and by closer inspection I found error in my proof.
    The error helped me also to find an obvious counterexample when ineqality doesn't hold.Consider the triangle with following parameters:
    [tex]a=b=1;c=\sqrt{3};R=1[/tex]
     
  9. Jan 23, 2007 #8
    I will rewrite the expression in a trigonometric form and give a restriction.

    [tex](sin(\alpha) sin(\beta) cosec(\gamma))^2+(sin(\alpha) cosec(\beta) sin(\gamma))^2+ (cosec(\alpha) sin(\beta) sin(\gamma))^2\geq\frac{9}{4}[/tex]

    The restriction is the inequality holds for acute triangles.Now,when I fixed it,
    proove the claim.
     
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