Question

If a, b, c are three sides of the \triangle ABC and A, B, C are their corresponding vertex, show that

a^2\tan\dfrac{A}{2}+b^2\tan\dfrac{B} {2}+c^2\tan\dfrac{C}{2}\geq 4S

in which S is the area of the triangle.

Collected in the board: Trigonometry

Steven Zheng posted 1 week ago

Answer

Using half angle identity and the law of cosines

\cos^2\dfrac{A}{2}=\dfrac{1+\cos A}{2}

=\dfrac{1+\dfrac{b^2+c^2-a^2}{2bc} }{2}

=\dfrac{(b+c)^2-a^2}{4bc}

=\dfrac{(b+c+a)(b+c-a)}{4bc}

\dfrac{a^2\tan\dfrac{A}{2}}{S} =\dfrac{a^2\cdotp \dfrac{\sin \dfrac{A}{2} }{\cos \dfrac{A}{2} } }{\dfrac{1}{2}bc\sin A } =\dfrac{a^2}{bc\cos^2\dfrac{A}{2} }

=\dfrac{4a^2}{(b+c+a)(b+c-a)}

Similarly, we can get,

\dfrac{b^2\tan\dfrac{B}{2} }{S}= \dfrac{4b^2}{(a+b+c)(a+c-b)}

\dfrac{c^2\cos\dfrac{C}{2} }{S}=\dfrac{4c^2}{(a+b+c)(a+b-c)}

On the other hand, the following inequality can be obtained by using Cauch Inequality.

(\dfrac{a^2}{b+c-a}+\dfrac{b^2}{a+c-b}+\dfrac{c^2}{a+b-c}) [(b+c-a)+(a+c-b)+(a+b-c)]\geq (a+b+c)^2

\dfrac{a^2}{b+c-a}+\dfrac{b^2}{a+c-b}+\dfrac{c^2}{a+b-c} \geq a+b+c

\dfrac{a^2\tan\dfrac{A}{2}}{S}+\dfrac{b^2\tan\dfrac{B}{2} }{S}+\dfrac{c^2\cos\dfrac{C}{2} }{S}

=(a+b+c)( \dfrac{a^2}{b+c-a}+\dfrac{b^2}{a+c-b}+\dfrac{c^2}{a+b-c})\geq 4

Therefore,

a^2\tan\dfrac{A}{2}+b^2\tan\dfrac{B} {2}+c^2\tan\dfrac{C}{2}\geq 4S

Steven Zheng posted 1 week ago

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