Let $ABC$ be a triangle. Prove that: $$\frac{1}{sinA}+\frac{1}{sinB}+\frac{1}{sinC}-\frac{1}{2}\left(tan\frac{A}{2}+tan\frac{B}{2}+tan\frac{C}{2}\right) \ge \sqrt{3} $$ My attempt: $$P=\frac{1}{sinA}+\frac{1}{sinB}+\frac{1}{sinC}-\frac{1}{2}\left(tan\frac{A}{2}+tan\frac{B}{2}+tan\frac{C}{2}\right) = \frac{1}{2}\left(cot\frac{A}{2}+cot\frac{B}{2}+cot\frac{C}{2}\right)$$ $$\alpha = cot\frac{A}{2}+cot\frac{B}{2}+cot\frac{C}{2}=cot\frac{A}{2}.cot\frac{B}{2}.cot\frac{C}{2}$$ $$\Leftrightarrow \alpha \ge3.\sqrt[3]{\alpha} \Leftrightarrow \alpha^2\ge27\Leftrightarrow \alpha\ge3\sqrt{3}$$ $$\Rightarrow P=\frac{1}{2}\alpha\ge\frac{3\sqrt{3}}{2}$$ Hmm where I was wrong ?
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1$\begingroup$ Use \ before writing a trig. function so that $cos(x)$ becomes $\cos(x)$. $\endgroup$ – vitamin d Feb 22 at 13:10
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You are not wrong because $$ \frac32\sqrt{3}\ge \sqrt3. $$
You just have proved a stronger inequality than required.
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Let $p$ is the semi-perimeter, $R$ is circumradius and $r$ is radius.
We have $$\frac{1}{\sin A}+\frac{1}{\sin B}+\frac{1}{\sin C} = \frac{2R}{a} + \frac{2R}{b} + \frac{2R}{c} = \frac{4R+r}{2p}+\frac{p}{2r},$$ and $$\tan\frac{A}{2}+\tan\frac{B}{2}+\tan\frac{C}{2} = \sum \frac{r}{b+c-a} = \frac{4R+r}{p}.$$ The inequality become $$p \geqslant 2\sqrt3 r.$$ Which is true by know inequality $$p \geqslant 3\sqrt 3 r.$$
