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Thread: Inequalities triangle

  1. #1
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    Inequalities triangle

    Prove that in any triangle ABC:
    $\sqrt{(\tan \frac{A}{2} + \tan \frac{B}{2})(\tan \frac{B}{2}+\tan \frac{C}{2})} + \sqrt{(\tan \frac{B}{2} + \tan \frac{C}{2})(\tan \frac{C}{2}+\tan \frac{A}{2})}+ \sqrt{(\tan \frac{C}{2} + \tan \frac{A}{2})(\tan \frac{A}{2}+\tan \frac{B}{2})} \leq 2(\cot A + \cot B + \cot C)$
    Last edited by leezangqe; Jan 26th 2013 at 03:10 AM.
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  2. #2
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    Re: Inequalities triangle

    Quote Originally Posted by leezangqe View Post
    Prove that in any triangle ABC:
    $\displaystyle \sqrt{(\tan \frac{A}{2} + \tan \frac{B}{2})(\tan \frac{B}{2}+\tan \frac{C}{2})} + \sqrt{(\tan \frac{B}{2} + \tan \frac{C}{2})(\tan \frac{C}{2}+\tan \frac{A}{2})}+ \sqrt{(\tan \frac{C}{2} + \tan \frac{A}{2})(\tan \frac{A}{2}+\tan \frac{B}{2})} \leq 2(\cot A + \cot B + \cot C)$
    .............
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    Re: Inequalities triangle

    Thank you help me latex, i try but i have this problem
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    Re: Inequalities triangle

    Quote Originally Posted by leezangqe View Post
    Prove that in any triangle ABC:
    $$\displaystyle \sqrt{(\tan \frac{A}{2} + \tan \frac{B}{2})(\tan \frac{B}{2}+\tan \frac{C}{2})} + \sqrt{(\tan \frac{B}{2} + \tan \frac{C}{2})(\tan \frac{C}{2}+\tan \frac{A}{2})}+ \sqrt{(\tan \frac{C}{2} + \tan \frac{A}{2})(\tan \frac{A}{2}+\tan \frac{B}{2})} \leq 2(\cot A + \cot B + \cot C)$$
    @leezangqe:select your whole latex code and click the summation symbol on the bar.
    we know that $\displaystyle A+B+B=180$
    lets start with the left side of the inequality:
    $\displaystyle \cot A+\cot B=\frac{\cos A}{\sin A}+\frac{\cos B}{\sin B}=\frac{\sin (A+B)}{\sin A \sin B}=\frac{2\sin C}{\cos (A-B)+\cos C}$
    (# as $\displaystyle \sin (A+B)=\sin C$ in a triangle, and using product-sum formula)
    now,$\displaystyle \frac{2\sin C}{\cos (A-B)+\cos C} \geq \frac{2\sin C}{1+\cos C}=2\tan \frac{C}{2}$
    so similarly we can write,
    $\displaystyle \cot A+\cot B \geq 2\tan \frac{C}{2}$
    $\displaystyle \cot B+\cot C \geq 2\tan \frac{A}{2}$
    $\displaystyle \cot C+\cot A \geq 2\tan \frac{B}{2}$
    Adding these together we get,
    $\displaystyle 2(\cot A + \cot B + \cot C) \geq 2(\tan \frac{A}{2}+\tan \frac{B}{2}+\tan \frac{C}{2}) ..................(i)$
    now lets simplify the right side:
    by the AM-GM inequality we can write:
    $\displaystyle \sqrt{(\tan \frac{A}{2} + \tan \frac{B}{2})(\tan \frac{B}{2}+\tan \frac{C}{2})} \geq \frac{\tan \frac{A}{2}+\tan \frac{B}{2}+\tan \frac{B}{2}+\tan \frac{C}{2}}{2}$
    or, similarly,
    $\displaystyle \sqrt{(\tan \frac{A}{2} + \tan \frac{B}{2})(\tan \frac{B}{2}+\tan \frac{C}{2})} \geq \frac{\tan \frac{A}{2}+2\tan \frac{B}{2}+\tan \frac{C}{2}}{2}$
    $\displaystyle \sqrt{(\tan \frac{B}{2} + \tan \frac{C}{2})(\tan \frac{C}{2}+\tan \frac{A}{2})} \geq \frac{\tan \frac{A}{2}+2\tan \frac{C}{2}+\tan \frac{B}{2}}{2}$
    $\displaystyle \sqrt{(\tan \frac{C}{2} + \tan \frac{A}{2})(\tan \frac{A}{2}+\tan \frac{B}{2})} \geq \frac{\tan \frac{B}{2}+2\tan \frac{A}{2}+\tan \frac{C}{2}}{2}$
    adding these we get,
    $\displaystyle \sqrt{(\tan \frac{A}{2} + \tan \frac{B}{2})(\tan \frac{B}{2}+\tan \frac{C}{2})} + \sqrt{(\tan \frac{B}{2} + \tan \frac{C}{2})(\tan \frac{C}{2}+\tan \frac{A}{2})}+ \sqrt{(\tan \frac{C}{2} + \tan \frac{A}{2})(\tan \frac{A}{2}+\tan \frac{B}{2})} \leq 2(\tan \frac{A}{2}+\tan \frac{B}{2}+\tan \frac{C}{2}) \leq 2(\cot A + \cot B + \cot C) $from $\displaystyle (i)$


    proved
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