Thread: Inequalities triangle

Prove that in any triangle ABC:
$\sqrt{(\tan \frac{A}{2} + \tan \frac{B}{2})(\tan \frac{B}{2}+\tan \frac{C}{2})} + \sqrt{(\tan \frac{B}{2} + \tan \frac{C}{2})(\tan \frac{C}{2}+\tan \frac{A}{2})}+ \sqrt{(\tan \frac{C}{2} + \tan \frac{A}{2})(\tan \frac{A}{2}+\tan \frac{B}{2})} \leq 2(\cot A + \cot B + \cot C)$

Originally Posted by leezangqe

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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Thank you help me latex, i try but i have this problem

Originally Posted by leezangqe

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