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Prove this inequality for all triangles

  1. Oct 16, 2012 #1
    1. The problem statement, all variables and given/known data
    Show that the angles a, b, c of each triangle satisfy this inequality.

    [itex]\tan \frac{a}{2}\tan \frac{b}{2} \tan \frac{c}{2} (\tan \frac{a}{2} + \tan \frac{b}{2} + \tan \frac{c}{2}) < \frac{1}{2}[/itex]


    2. Relevant equations
    3. The attempt at a solution

    I used the half angle formula: [itex] \tan \frac{\theta}{2} = \frac{1- \cos x}{\sin x} [/itex]

    That gives

    [tex]\frac{(1- \cos a)(1- \cos b)(1- \cos c)}{(\sin a)(\sin b)(\sin c)} [\frac{1- \cos a}{\sin a} + \frac{1- \cos b}{\sin b} + \frac{1- \cos c}{\sin c}] < \frac{1}{2}[/tex]

    I don't know where to go from this. Can someone give me any hints please? Thanks.
     
  2. jcsd
  3. Oct 17, 2012 #2

    SammyS

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    One hint is that a + b + c = π .
     
  4. Oct 18, 2012 #3
    Thanks SammyS.

    Since a + b + c = π we can prove that (cot A)(cot B)(cot C) = cot A + cot B + cot C
    With that I did the following.

    Rearrange given equation:
    2[tan(a/2) + tan(b/2) + tan(c/2)] < [tan(a/2) * tan(b/2) * tan(c/2)]^-1

    2[tan(a/2) + tan(b/2) + tan(c/2)] < cot(a/2) * cot(b/2) * cot (c/2)

    Substitute RHS

    2[tan(a/2) + tan(b/2) + tan(c/2)] < cot(a/2) + cot(b/2) + cot (c/2)

    From here on, everything I try leads to no where. I tried expressing c/2 as (pi/2)-(a+b) but that makes the problem even more complicated. I honestly give up. Can someone please help me? I suck at this and I need to solve this problem for tomorrow. I'm desperate. Thanks.
     
  5. Oct 18, 2012 #4
    Can someone please please help me? I only have about two more hours to solve this stupid question.
     
  6. Oct 18, 2012 #5
    No help at all? :(
     
  7. Oct 18, 2012 #6

    SammyS

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    I'm sorry, but I don't have the solution either.

    My thought with a + b + c = π was simply that then

    [itex]\displaystyle c=\pi-(a+b)\ \ \text{ so that }\ \ \frac{c}{2}=\frac{\pi}{2}-\frac{a+b}{2}\,,[/itex]

    thus [itex]\displaystyle \tan\left(\frac{c}{2}\right)=\tan\left(\frac{\pi}{2}-\frac{a+b}{2}\right)=\cot\left(\frac{a+b}{2}\right)\ .[/itex]

    I haven't figured out if that leads anywhere.


    How did you conclude the following:
    Since a + b + c = π we can prove that (cot A)(cot B)(cot C) = cot A + cot B + cot C​
    ?
     
  8. Oct 18, 2012 #7
    http://in.answers.yahoo.com/question/index?qid=20080111225452AADKNLv

    I was told that

    tan(a/2)tan(b/2)+tan(a/2)tan(c/2)+tan(b/2)tan(c/2) =1 is useful.

    You can prove that by taking the tan of both sides of the following

    (a+b)/2 = 90 -c

    Thanks for trying.
     
  9. Oct 18, 2012 #8

    SammyS

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    I looked that over & I'm quite sure there is a sign error in that derivation.
     
  10. Oct 20, 2012 #9

    ehild

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    I know it is too late, but I think I found the solution just now.

    As SammyS pointed out,
    [tex]\tan(\frac{c}{2})=\cot(\frac{a+b}{2})[/tex]

    Expanding: [tex]\cot(\frac{a+b}{2})=\frac{1-\tan(\frac{a}{2})\tan(\frac{b}{2})}{\tan(\frac{a}{2})+\tan(\frac{b}{2})}[/tex]

    With the notations x=tan(a/2) and y=tan(b/2), (x≥0, y≥0) the original expression becomes

    [tex]A=xy \frac{1-xy}{x+y} \left( x+y+\frac{1-xy}{x+y} \right)=xy\left(1-xy+\frac{(1-xy)^2}{(x+y)^2}\right)[/tex]

    From the relation between the arithmetic and geometric means (x+y)2≥4xy, so
    [tex]\frac{(1-xy)^2}{(x+y)^2} ≤ \frac{(1-xy)^2}{4xy}[/tex]

    [tex]A≤xy\left(1-xy+\frac{(1-xy)^2}{4xy}\right)[/tex]

    denoting xy by z (z≥0):

    [tex]A≤z\left(1-z+\frac{(1-z)^2}{4z}\right)=\frac{1}{4}(-3z^2+2z+1)[/tex], which is an upside-down parabola.

    Check, please.

    ehild
     
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