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Triangle Inequalities.

  1. Oct 15, 2008 #1
    1. The problem statement, all variables and given/known data
    a,b,c are the sides of the triangle. then find the range of [tex]\frac {a^{2}+b^{2}+c^{2}}{ab+bc+ca}[/tex]


    3. The attempt at a solution
    Let;
    [tex]\frac {a^{2}+b^{2}+c^{2}}{ab+bc+ca}=k[/tex]
    cross multiplying and adding 2(ab+bc+ca) on both sides
    then;

    [tex]\frac{(a+b+c)^{2}}{ab+bc+ca}=k+2[/tex]

    Applying sin rule:
    a=2R sin A
    b=2R sin B
    c=2R sin C
    R is the circumradius.
    now k+2=[tex]\frac{(sin A+ sin B+ sin C)^{2}}{sin A sin B+sin B sin C+ sin C sin A}[/tex]
    This can be further modified as:
    k+2= [tex]\frac{(sin A+ sin B+ sin C)^{2}}{sin A sin B sin C (cosec A+Cosec B+cosec C)}[/tex]
    As A+B+C= 180
    I can say that maximum value of [tex]sin A+sin B+sin C= \frac{3^{1.5}}{2}[/tex]

    I also know that the minimum value of [tex]cosec A+ cosec B+ cosec C=2\sqrt{3}[/tex]
    Both these inequalities hold when triangle is equilateral. But I am stuck with the expression sinA. Sin B. sin C??? Am I on the right track???
     
  2. jcsd
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